# What is the intuition behind beta distribution?

Disclaimer: I'm not a statistician but a software engineer. Most of my knowledge in statistics comes from self-education, thus I still have many gaps in understanding concepts that may seem trivial for other people here. So I would be very thankful if answers included less specific terms and more explanation. Imagine that you are talking to your grandma :)

I'm trying to grasp the nature of beta distribution – what it should be used for and how to interpret it in each case. If we were talking about, say, normal distribution, one could describe it as arrival time of a train: most frequently it arrives just in time, a bit less frequently it is 1 minute earlier or 1 minute late and very rarely it arrives with difference of 20 minutes from the mean. Uniform distribution describes, in particular, chance of each ticket in lottery. Binomial distribution may be described with coin flips and so on. But is there such intuitive explanation of beta distribution?

Let's say, $$\alpha=.99$$ and $$\beta=.5$$. Beta distribution $$B(\alpha, \beta)$$ in this case looks like this (generated in R): But what does it actually mean? Y-axis is obviously a probability density, but what is on the X-axis?

I would highly appreciate any explanation, either with this example or any other.

• The y-axis is not a probability (which is obvious, because by definition a probability cannot lie outside the interval $[0,1]$, but this plot extends up to $50$ and--in principle--to $\infty$). It is a probability density: a probability per unit of $x$ (and you have described $x$ as a rate).
– whuber
Jan 15, 2013 at 16:40
• @whuber: yeah, I understand what PDF is - that was just mistake in my description. Thanks for a valid note! Jan 15, 2013 at 19:32
• I'll try and find the reference but I know some of the more bizarre shapes for the generalized Beta distribution with form $a + (b-a)Beta(\alpha_1,\alpha_2)$ have applications such as physics. Also, you can fit it to expert data (min, mode, max) in data-poor environments and it is often better than using a Triangular distribution (unfortunately often used by IEs). Sep 20, 2018 at 1:40
• You've obviously never traveled with the railway company Deutsche Bahn. You'd be less optimistic. Apr 19, 2019 at 11:06

The short version is that the Beta distribution can be understood as representing a distribution of probabilities, that is, it represents all the possible values of a probability when we don't know what that probability is. Here is my favorite intuitive explanation of this:

Anyone who follows baseball is familiar with batting averages—simply the number of times a player gets a base hit divided by the number of times he goes up at bat (so it's just a percentage between 0 and 1). .266 is in general considered an average batting average, while .300 is considered an excellent one.

Imagine we have a baseball player, and we want to predict what his season-long batting average will be. You might say we can just use his batting average so far- but this will be a very poor measure at the start of a season! If a player goes up to bat once and gets a single, his batting average is briefly 1.000, while if he strikes out, his batting average is 0.000. It doesn't get much better if you go up to bat five or six times- you could get a lucky streak and get an average of 1.000, or an unlucky streak and get an average of 0, neither of which are a remotely good predictor of how you will bat that season.

Why is your batting average in the first few hits not a good predictor of your eventual batting average? When a player's first at-bat is a strikeout, why does no one predict that he'll never get a hit all season? Because we're going in with prior expectations. We know that in history, most batting averages over a season have hovered between something like .215 and .360, with some extremely rare exceptions on either side. We know that if a player gets a few strikeouts in a row at the start, that might indicate he'll end up a bit worse than average, but we know he probably won't deviate from that range.

Given our batting average problem, which can be represented with a binomial distribution (a series of successes and failures), the best way to represent these prior expectations (what we in statistics just call a prior) is with the Beta distribution- it's saying, before we've seen the player take his first swing, what we roughly expect his batting average to be. The domain of the Beta distribution is (0, 1), just like a probability, so we already know we're on the right track, but the appropriateness of the Beta for this task goes far beyond that.

We expect that the player's season-long batting average will be most likely around .27, but that it could reasonably range from .21 to .35. This can be represented with a Beta distribution with parameters $$\alpha=81$$ and $$\beta=219$$:

curve(dbeta(x, 81, 219)) I came up with these parameters for two reasons:

• The mean is $$\frac{\alpha}{\alpha+\beta}=\frac{81}{81+219}=.270$$
• As you can see in the plot, this distribution lies almost entirely within (.2, .35)- the reasonable range for a batting average.

You asked what the x axis represents in a beta distribution density plot—here it represents his batting average. Thus notice that in this case, not only is the y-axis a probability (or more precisely a probability density), but the x-axis is as well (batting average is just a probability of a hit, after all)! The Beta distribution is representing a probability distribution of probabilities.

But here's why the Beta distribution is so appropriate. Imagine the player gets a single hit. His record for the season is now 1 hit; 1 at bat. We have to then update our probabilities- we want to shift this entire curve over just a bit to reflect our new information. While the math for proving this is a bit involved (it's shown here), the result is very simple. The new Beta distribution will be:

$$\mbox{Beta}(\alpha_0+\mbox{hits}, \beta_0+\mbox{misses})$$

Where $$\alpha_0$$ and $$\beta_0$$ are the parameters we started with- that is, 81 and 219. Thus, in this case, $$\alpha$$ has increased by 1 (his one hit), while $$\beta$$ has not increased at all (no misses yet). That means our new distribution is $$\mbox{Beta}(81+1, 219)$$, or:

curve(dbeta(x, 82, 219)) Notice that it has barely changed at all- the change is indeed invisible to the naked eye! (That's because one hit doesn't really mean anything).

However, the more the player hits over the course of the season, the more the curve will shift to accommodate the new evidence, and furthermore the more it will narrow based on the fact that we have more proof. Let's say halfway through the season he has been up to bat 300 times, hitting 100 out of those times. The new distribution would be $$\mbox{Beta}(81+100, 219+200)$$, or:

curve(dbeta(x, 81+100, 219+200)) Notice the curve is now both thinner and shifted to the right (higher batting average) than it used to be- we have a better sense of what the player's batting average is.

One of the most interesting outputs of this formula is the expected value of the resulting Beta distribution, which is basically your new estimate. Recall that the expected value of the Beta distribution is $$\frac{\alpha}{\alpha+\beta}$$. Thus, after 100 hits of 300 real at-bats, the expected value of the new Beta distribution is $$\frac{81+100}{81+100+219+200}=.303$$- notice that it is lower than the naive estimate of $$\frac{100}{100+200}=.333$$, but higher than the estimate you started the season with ($$\frac{81}{81+219}=.270$$). You might notice that this formula is equivalent to adding a "head start" to the number of hits and non-hits of a player- you're saying "start him off in the season with 81 hits and 219 non hits on his record").

Thus, the Beta distribution is best for representing a probabilistic distribution of probabilities: the case where we don't know what a probability is in advance, but we have some reasonable guesses.

• Here's a similar example from John Cook using binary Amazon seller rankings with different number of reviews. The discussion of choosing a prior in the comments is particularly illuminating: johndcook.com/blog/2011/09/27/bayesian-amazon/#comments Jan 15, 2013 at 19:58
• You should point out that the prior need not be beta-distributed (unless you go with the Jeffreys' prior, $\alpha_0=\beta_0=1/2$ — only the likelihood must be beta distributed. Jan 16, 2013 at 18:08
• + I like your explanation of how you update the distribution when you have more data. Jan 16, 2013 at 21:44
• @user27997 Those gave the desired mean of .27, and a standard deviation that is very roughly realistic for batting averages (about .025). Incidentally, I give an explanation of how to calculate α and β from a desired mean and variance here. Jul 13, 2013 at 23:38
• Hi @DavidRobinson, how did you decide $\alpha_0$ and $\beta_0$ to be 81 and 239, but not something else but with the same ratio (like 40, 115)? Is it because you're assuming a player plays (hits or misses) 300 times in a season? Dec 5, 2017 at 19:45

A Beta distribution is used to model things that have a limited range, like 0 to 1.

Examples are the probability of success in an experiment having only two outcomes, like success and failure. If you do a limited number of experiments, and some are successful, you can represent what that tells you by a beta distribution.

Another example is order statistics. For example, if you generate several (say 4) uniform 0,1 random numbers, and sort them, what is the distribution of the 3rd one?

I use them to understand software performance diagnosis by sampling. If you stop a program at random $n$ times, and $s$ of those times you see it doing something you could actually get rid of, and $s>1$, then the fraction of time to be saved by doing so is represented by $Beta(s+1, (n-s)+1)$, and the speedup factor has a BetaPrime distribution.

The Beta distribution also appears as an order statistic for a random sample of independent uniform distributions on $(0,1)$.

Precisely, let $U_1$, $\ldots$, $U_n$ be $n$ independent random variables, each having the uniform distribution on $(0,1)$. Denote by $U_{(1)}$, $\ldots$, $U_{(n)}$ the order statistics of the random sample $(U_1, \ldots, U_n)$, defined by sorting the values of $U_1$, $\ldots$, $U_n$ in increasing order. In particular $U_{(1)}=\min(U_i)$ and $U_{(n)}=\max(U_i)$. Then one can show that $U_{(k)} \sim \textrm{Beta}(k, n+1-k)$ for every $k=1,\ldots,n$.

This result shows that the Beta distributions naturally appear in mathematics, and it has some interesting applications in mathematics.

There are two principal motivations:

First, the beta distribution is conjugate prior to the Bernoulli distribution. That means that if you have an unknown probability like the bias of a coin that you are estimating by repeated coin flips, then the likelihood induced on the unknown bias by a sequence of coin flips is beta-distributed.

Second, a consequence of the beta distribution being an exponential family is that it is the maximum entropy distribution for a set of sufficient statistics. In the beta distribution's case these statistics are $\log(x)$ and $\log(1-x)$ for $x$ in $[0,1]$. That means that if you only keep the average measurement of these sufficient statistics for a set of samples $x_1, \dots, x_n$, the minimum assumption you can make about the distribution of the samples is that it is beta-distributed.

The beta distribution is not special for generally modeling things over [0,1] since many distributions can be truncated to that support and are more applicable in many cases. Let's assume a seller on some e-commerce web-site receives 500 ratings of which 400 are good and 100 are bad.

We think of this as the result of a Bernoulli experiment of length 500 which led to 400 successes (1 = good) while the underlying probability $p$ is unknown.

The naive quality in terms of ratings of the seller is 80% because 0.8 = 400 / 500. But the "true" quality in terms of ratings we don't know.

Theoretically also a seller with "true" quality of $p=77\%$ might have ended up with 400 good of 500 ratings.

The pointy bar plot in the picture represents the frequency of how often it happend in a simulation that for a given assumed "true" $p$ 400 of 500 ratings were good. The bar plot is the density of the histogram of the result of the simulation.

And as you can see - the density curve of the beta distribution for $\alpha=400+1$ and $\beta=100+1$ (orange) tightly surrounds the bar chart (the density of the histogram for the simulation).

So the beta distribution essentially defines the probability that a Bernoulli experiment's success probability is $p$ given the outcome of the experiment.

library(ggplot2)

# 90% positive of 10 ratings
o1 <- 9
o0 <- 1
M <- 100
N <- 100000

m <- sapply(0:M/M,function(prob)rbinom(N,o1+o0,prob))
v <- colSums(m==o1)
df_sim1 <- data.frame(p=rep(0:M/M,v))
df_beta1 <- data.frame(p=0:M/M, y=dbeta(0:M/M,o1+1,o0+1))

# 80% positive of 500 ratings
o1 <- 400
o0 <- 100
M <- 100
N <- 100000

m <- sapply(0:M/M,function(prob)rbinom(N,o1+o0,prob))
v <- colSums(m==o1)
df_sim2 <- data.frame(p=rep(0:M/M,v))
df_beta2 <- data.frame(p=0:M/M, y=dbeta(0:M/M,o1+1,o0+1))

ggplot(data=df_sim1,aes(p)) +
scale_x_continuous(breaks=0:10/10) +

geom_histogram(aes(y=..density..,fill=..density..),
binwidth=0.01, origin=-.005, colour=I("gray")) +
geom_line(data=df_beta1 ,aes(p,y),colour=I("red"),size=2,alpha=.5) +

geom_histogram(data=df_sim2, aes(y=..density..,fill=..density..),
binwidth=0.01, origin=-.005, colour=I("gray")) +
geom_line(data=df_beta2,aes(p,y),colour=I("orange"),size=2,alpha=.5)


http://www.joyofdata.de/blog/an-intuitive-interpretation-of-the-beta-distribution/

• Thank you for your contribution! I am puzzled about something, though: although the histogram legend states they show beta densities, you appear to claim these also describe the outcomes of binomial simulations ("how often it happend in a simulation"). But the two are different things, even though they happen to appear fairly close in the illustration. (That's a consequence of the near-normality of the Beta with large parameters and the Central Limit theorem for Binomial distributions.)
– whuber
Nov 15, 2013 at 21:12
• That is a good point! But I am not sure how to rephrase it propperly. If I would just plot the histogram then, of course, you wouldn't see much of the density given the magnitude of it. So yes, the histogram is actually I guess not just scaled down but actually the (estimated) density of the original histogram. Given the number of runs I could also figure out a factor and scale it down linearly but it would look almost exactly the same PLUS what I (actually) want to compare is the density of beta with the density of the result of the simulation (the density of the original histogram). Nov 15, 2013 at 21:33

So far the preponderance of answers covered the rationale for Beta RVs being generated as the prior for a sample proportions, and one clever answer has related Beta RVs to order statistics.

Beta distributions also arise from a simple relationship between two Gamma(k_i, 1) RVs, i=1,2 call them X and Y. X/(X+Y) has a Beta distribution.

Gamma RVs already have their rationale in modeling arrival times for independent events, so I will not address that since it is not your question. But a "fraction of time" spent completing one of two tasks performed in sequence naturally lends itself to a Beta distribution.

• +1 Thanks for pointing that out about using Gamma to form a Beta distribution. I've heard that if you want to generalize the Beta into a Dirichlet, you simply put more Gammas in the denominator. Maybe a statistician just knows that, but to me that was really useful when looking at confidence intervals of a categorical observation. Mar 23, 2016 at 15:30

Most of the answers here seem to cover two approaches: Bayesian and the order statistic. I'd like to add a viewpoint from the binomial, which I think the easiest to grasp.

The intuition for a beta distribution comes into play when we look at it from the lens of the binomial distribution. The difference between the binomial and the beta is that the former models the number of occurrences ($$x$$), while the latter models the probability ($$p$$) itself. In other words, the probability is a parameter in binomial; In the Beta, the probability is a random variable.

## Interpretation of $$\boldsymbol{\alpha}$$, $$\boldsymbol{\beta}$$

You can think of $$\alpha-1$$ as the number of successes and $$\beta-1$$ as the number of failures, just like $$n$$ & $$n-x$$ terms in binomial. You can choose the $$\alpha$$ and $$\beta$$ parameters however you think they are supposed to be. If you think the probability of success is very high, let's say 90%, set 90 for $$\alpha$$ and 10 for $$\beta$$. If you think otherwise, 90 for $$\beta$$ and 10 for $$\alpha$$.

As $$\alpha$$ becomes larger (more successful events), the bulk of the probability distribution will shift towards the right, whereas an increase in $$\beta$$ moves the distribution towards the left (more failures). Also, the distribution will narrow if both $$\alpha$$ and $$\beta$$ increase, for we are more certain.

## The Intuition behind the shapes

The PDF of Beta distribution can be U-shaped with asymptotic ends, bell-shaped, strictly increasing/decreasing or even straight lines. As you change $$\alpha$$ or $$\beta$$, the shape of the distribution changes.

### a. Bell-shape Notice that the graph of PDF with $$\alpha = 8$$ and $$\beta = 2$$ is in blue, not in read. The x-axis is the probability of success. The PDF of a beta distribution is approximately normal if $$\alpha +\beta$$ is large enough and $$\alpha$$ & $$\beta$$ are approximately equal.

### b. Straight Lines The beta PDF can be a straight line too.

### c. U-shape When $$\alpha <1$$, $$\beta<1$$, the PDF of the Beta is U-shaped.

### The Intuition behind the shapes

Why would Beta(2,2) be bell-shaped?

If you think of $$\alpha-1$$ as the number of successes and $$\beta-1$$ as the number of failures, Beta(2,2) means you got 1 success and 1 failure. So it makes sense that the probability of the success is highest at 0.5.

Also, Beta(1,1) would mean you got zero for the head and zero for the tail. Then, your guess about the probability of success should be the same throughout [0,1]. The horizontal straight line confirms it.

What’s the intuition for Beta(0.5, 0.5)?

Why is it U-shaped? What does it mean to have negative (-0.5) heads and tails? I don’t have an answer for this one yet. I even asked this on Stackexchange but haven’t gotten the response yet. If you have a good idea about the U-shaped Beta, please let me know!

• What do you mean by 'models the probability'? And what do you mean by 'the probability is a random variable' outside the Bayesian context? Jan 8, 2020 at 21:16
• I really like this approach, they two are naturally related. The binomial coefficient is defined as a product of three factorials. The p.d.f.'s normalizing Beta function can be defined by a product of three Gamma functions. A Gamma function is literally a hack (okay, 'analytic continuation', whatever) for extending factorials from positive integers to all numbers. In fact, a Beta where the Gammas' parameters are integers gives you the same value as a binomial coefficient with k+1 and vice versa.
– jkm
Jan 8, 2020 at 21:27
• This answer seems intuitive, but it's a bit misleading. I agree that the beta distribution can model a density over probabilities $p \in [0, 1]$. However, there is no reason to prefer that parametrization. For example, you could just as easily model a density over the odds $\frac{p}{1-p}$, which is called the beta-prime distribution. You could model the log-odds, or you scale the probability $p$ [0, 1] to $[0, \frac{\pi}{2}]$, and model a density over the sin of that number. In each case, the shape of the density is different. Jun 10, 2021 at 13:21
• Although your reasoning about the maximum likelihood estimate is correct, the part of your answer that's misleading is when you say "you got zero for the head and zero for the tail. Then, your guess about the probability of success should be the same throughout [0,1]." This assumes a uniform prior, and that the uniform distribution on [0, 1] is minimally assumptive. There's no reason to believe that. If you had arbitrarily chosen a different parametrization, your uniform distribution would correspond to a different $\alpha, \beta$. Incidentally, an invariant prior distribution would be Jun 10, 2021 at 13:24
• ...the Jeffreys prior, which corresponds to the arcsine distribution. Equivalently, this is the uniform distribution if you had chosen the odd-seeming parametrization using the sine of the angle I described above. Jun 10, 2021 at 13:26

My intuition says that it "weighs" both the current proportion of success "$x$" and current proportion of failure "$(1-x)$": $f(x;\alpha,\beta) = \text{constant}\cdot x^{\alpha-1}(1-x)^{\beta-1}$. Where the constant is $1/B(\alpha,\beta)$. The $\alpha$ is like a "weight" for success's contribution. The $\beta$ is like a "weight" for failure's contribution. You have a two dimensional parameter space (one for successes contribution and one for failures contribution) which makes it kind of difficult to think about and understand.

In the cited example the parameters are alpha = 81 and beta = 219 from the prior year [81 hits in 300 at bats or (81 and 300 - 81 = 219)]

I don't know what they call the prior assumption of 81 hits and 219 outs but in English, that's the a priori assumption.

Notice how as the season progresses the curve shifts left or right and the modal probability shifts left or right but there is still a curve.

I wonder if the Laa of Large Numbers eventually takes hold and drives the batting average back to .270.

To guesstimate the alpha and beta in general one would take the complete number of prior occurrences (at bats), the batting average as known, obtain the total hits (the alpha), the beta or the grand total minus the failures) and voila – you have your formula. Then, work the additional data in as shown.

• > In the cited example the parameters are alpha = 81 and beta = 219 from the prior year [81 hits in 300 at bats or (81 and 300 - 81 = 219)] Are we sure about that? What I am not seeing in the Robinson explanation here is 'what initial prior (before any 'at bats') is combined with 81 hits in 300 'at bats' to get the posterior Beta(81, 219)? Mar 28 at 18:29

The beta distribution is very useful when you are working with particle size distribution. It is not the situation when you want to model a grain distribution; this case is better to use Tanh distribution $F(X) = \tanh ((x/p)^n)$ that is not bounded on the right.

By the way, what's up if you produce a size distribution from a microscopic observation and you have a particle distribution in number, and your aim is to work with a volume distribution? It is almost mandatory to get the original distribution in number bounded on the right. So, the transformation is more consistent because you are sure that in the new volume distribution does not appear any mode, nor median nor medium size out of the interval you are working. Besides, you avoid the Greenland Africa effect.

The transformation is very easy if you have regular shapes, i.e., a sphere or a prism. You ought to add three units to the alpha parameter of the number beta distribution and get the volume distribution.

• Welcome to the site. Was this intended as an answer to the OP's question? Can you clarify how this relates to the intuition behind the beta distribution? Nov 4, 2015 at 20:38
• Please edit to clarify the intuition about a beta distribution. Nov 4, 2015 at 21:58

In another question concerning the beta distribution the following intuition behind beta is provided:

In other words the beta distribution can be seen as the distribution of probabilities in the center of a jittered distribution.

If you break a unit-length rod into k+m pieces, keeping k and discarding m, then the resulting length is Beta(k,m).

(See this question for more details. A related example is that Beta(k,n-k) is the k-th smallest among n-1 independent variables uniformly distributed over the unit interval.)

There are already so many awesome answers here, but I'd like to share with you how I interpret the "probabilistic distribution of probabilities" as @David Robinson described in the accepted answer and add some supplementary points using some very simple illustrations and derivations.

Imagine this, we have a coin and flip it in the following three scenarios: 1) toss it five times and get TTTTT (five tails and zero head); in scenario 2) use the same coin and toss it also five times and get HTTHH (three heads and two tails); in scenario 3) get the same coin and toss it ten times and get THHTHHTHTH (six heads and four tails).

Then three issues arise a) we don't have a strategy to guess the probability in the first flipping; b) in scenario 1 the probability (we would work out) of getting head in the 6th tossing would be impossible which seems unreal(black swan event); c) in scenario 2 and 3 the (relative) probabilities of getting head next time are both $$0.6$$ although we know the confidence is higher in scenario 3. Therefore it is not enough to estimate the probability in tossing a coin just using a probability point and with no prior information, instead, we need a prior before we toss the coin and a probability distribution for each time step in the three cases above.

Beta distribution $$\text{Beta}(\theta|\alpha_H, \alpha_T)$$ can address the three problems where $$\theta$$ represents the density over the interval [0, 1], $$\alpha_H$$ the times heads occur and $$\alpha_T$$ the times tails occur here.

For the issue a, we can assume before flipping the coin that heads and tails are equally likely by either use a probability point and saying that the chance of occurring heads is 50%, or employing the Beta distribution and setting the prior as $$\text{Beta}(\theta|1, 1)$$ (equivalent to the uniform distribution) meaning two virtual tosses(we can treat the hyperparameter (1, 1) as pseudocounts) and we have observed one head event and one tail event (as depicted bellow).

p = seq(0,1, length=100)
plot(p, dbeta(p, 1, 1), ylab="dbeta(p, 1, 1)", type ="l", col="blue") In fact we can bridge the two methods by the following derivation:

\begin{align*} E[\text{Beta}(\theta|\alpha_H, \alpha_T)] &= \int_0^1 \theta P(\theta|\alpha_H, \alpha_T) d\theta \hspace{2.15cm}\text{the numerator/normalization is a constant}\\ &=\dfrac{\int_0^1 \theta \{ \theta^{\alpha_H-1} (1-\theta)^{\alpha_T-1}\}\ d\theta}{B(\alpha_H,\alpha_T)}\hspace{.75cm} \text{definition of Beta; the numerator is a constant} \\ &= \dfrac{B(\alpha_H+1,\alpha_T)}{B(\alpha_H,\alpha_T)} \hspace{3cm}\text{\theta \theta^{\alpha_H-1}=\theta^{\alpha_H}} \\ &= \dfrac{\Gamma(\alpha_H+1) \Gamma(\alpha_T)}{\Gamma(\alpha_H+\alpha_T+1)} \dfrac{\Gamma(\alpha_H+\alpha_T)}{\Gamma(\alpha_H)\Gamma(\alpha_T)} \\ &= \dfrac{\alpha_H}{\alpha_H+\alpha_T} \end{align*}

We see that the expectation $$\frac{1}{1+1}=50%$$ is just equal to the probability point, and we can also view the probability point as one point in the Beta distribution (the Beta distribution implies that all probabilities are 100% but the probability point implies that only 50% is 100%).

For the issue b, we can calculate the posterior as follows after getting N observations(N is 5: $$N_T=5$$ and $$N_H=0$$) $$\mathcal{D}$$.

\begin{align*} \text{Beta}(\theta|\mathcal{D}, \alpha_H, \alpha_T) &\propto P(\mathcal{D}|\theta,\alpha_H, \alpha_T)P(\theta|\alpha_H, \alpha_T) \hspace{.47cm}\text{likelihood \times prior}\\ &= P(\mathcal{D}|\theta) P(\theta|\alpha_H, \alpha_T) \hspace{2cm} \text{as depicted bellow}\\ &\propto \theta^{N_H} (1-\theta)^{N_T} \cdot \theta^{\alpha_H-1} (1-\theta)^{\alpha_T-1} \\ &= \theta^{N_H+\alpha_H-1} (1-\theta)^{N_T+\alpha_T-1} \\ &= \text{Beta}(\theta|\alpha_H+N_H, \alpha_T+N_T) \end{align*} $$\mathcal{D}$$,$$\alpha_H$$ and $$\alpha_T$$ are independent given $$\theta$$

We can plug in the prior and N observations and get $$\text{Beta}(\theta|1+0, 1+5)$$

p = seq(0,1, length=100)
plot(p, dbeta(p, 1+0, 1+5), ylab="dbeta(p, 1+0, 1+5)", type ="l", col="blue") We see the distribution over all probabilities of getting a head the density is high over the low probabilities but never be zero we can get otherwise, and the expectation is $$E[\text{Beta}(\theta|1+0, 1+5)] = \frac{1+0}{1+0+1+5}$$ (the Laplace smoothing or additive smoothing) rather than 0/impossible (in issue b).

For the issue c, we can calculate the two posteriors (along the same line as the above derivation) and compare them (as with the uniform as prior). When we get three heads and two tails we get $$\text{Beta}(\theta|\mathcal{D}, \alpha_H, \alpha_T)=\text{Beta}(\theta|1+3, 1+2)$$

p = seq(0,1, length=100)
plot(p, dbeta(p, 1+3, 1+2), ylab="dbeta(p, 1+3, 1+2)", type ="l", col="blue") When we get six heads and four tails we get $$\text{Beta}(\theta|\mathcal{D}, \alpha_H, \alpha_T)=\text{Beta}(\theta|1+6, 1+4)$$

p = seq(0,1, length=100)
plot(p, dbeta(p, 1+6, 1+4), ylab="dbeta(p, 1+6, 1+4)", type ="l", col="blue") We can calculate their expectations ($$\frac{1+3}{1+3+1+2} = 0.571 \approx \frac{1+6}{1+6+1+4} = 0.583$$, and if we don't consider the prior $$\frac{3}{3+2} = \frac{6}{6+4}$$) but we can see that the second curve is more tall and narrow(more confident). The denominator of the expectation can be interpreted as a measure of confidence, the more evidence (either virtual or real) we have the more confident the posterior and the taller and narrower the curve of the Beta distribution. But if we do like that in issue c the information is just lost.

References:

To add to David Robinson's answer, the initial $$\alpha$$ and $$\beta$$ parameters of the Beta distribution can be computed from a desired mean ($$\mu$$) and standard deviation ($$\sigma$$) using the following formulae:

$$\alpha = \frac{-\mu(\mu^2-\mu+\sigma^2)}{\sigma^2}$$

$$\beta = \frac{(\mu-1)(\mu^2-\mu+\sigma^2)}{\sigma^2}$$

For example, if the desired mean = 0.27 and standard deviation = 0.025, then:

$$\alpha = \frac{-0.27(0.27^2-0.27+0.025^2)}{0.025^2} \approx 84.88$$

$$\beta = \frac{(0.27-1)(0.27^2-0.27+0.025^2)}{0.025^2} \approx 229.48$$

Compare these to the estimates in David's answer ($$\alpha=81, \beta=219$$)

Based on the mean and variance (table on the right) of the Beta distribution and solved via WolframAlpha here.

• In any two-parameter family where the SD is determined by the mean--of which there are a huge number (such as all location-scale families)--a result like this will hold: the parameters determine the mean and SD and vice versa . This result reveals something about algebra but nothing about statistics.
– whuber
Sep 30, 2021 at 17:32

I think there is NO intuition behind beta distribution! The beta distribution is just a very flexible distribution with FIX range! And for integer a and b it is even easy to deal with. Also many special cases of the beta have their native meaning, like the uniform distribution. So if the data needs to be modeled like this, or with slightly more flexibility, then the beta is a very good choice.

• I disagree, but I have seen multiple external sources espousing this claim that no cute intuitive real-world pedagogical example exists for a beta distribution. Aug 10, 2020 at 2:21
• I believe there is certainly an opportunity to understand the distribution intuitively, as many of the above answers do, however, in practice, this answer isn't entirely incorrect either. Many times in practical math we add in terms (e.g. log, exponent, etc) in order to try to fit a theoretical graph onto noisy, practical data. You can then try to reason why those specific terms work, but in practice, we were just trying stuff out to fit our observation.
– ZAR
Dec 16, 2021 at 15:49