New answers tagged probability
0
The empirical rule applies when the observations appear to be drawn from a population that is normal. That population could be standard normal or $N(-57632, 838074)$. It’s just the normality that matters.
1
Interpretation 2: No red and blue balls within the same 1m x 1m square of a grid
In this case, the problem is similar to an urn model: you have $100^2$ urns in which you place red and blue balls, and you question yourself whether any red and blue are in the same urn.
Computation
We can compute how likely it is that the 10 blue balls occupy $k$ urns, which ...
0
let's presume that you have 10000 boxes, and you place blue balls in these boxes at random, and then you are placing red balls at random, let's count probability of success:
$$p_b(i) = p(\text{10 blue balls in i boxes}) = \frac{\binom{10000}{i}\binom{9}{i-1}}{10000^{10}}$$
$$p_r(i) = p(\text{100 red balls avoid i boxes with blue balls}) = (\frac{10000-i}{...
0
You can calculate the probability $P$ of you being selected by using $P = 1-\bar{P}$, where $\bar{P}$ is the probability that you are not being selected.
For the first round, we have $\bar{P}_1 = \frac{54}{56}$. The second round is more complicated, as we don't know the number of entries remaining. I don't see any other way than doing this the hard way:
With ...
1
The Empirical Rule with which I am familiar looks at sample means and standard deviations. (For example, the interval $\bar X \pm S$ often contains about 68% of the sample elements.)
Boxplots look at medians and interquartile ranges.
Of course, a boxplot is constructed so that (nearly) 50% of the data is 'contained' within the box.
The ER is often said to ...
2
Yes, it is. As you mentioned, the classical rule is $P(A,B) = P(A|B)P(B)$, but it can also be applied to conditional probabilities like $P(\cdot|C)$ instead of $P(\cdot)$. It then becomes
$$
P(A,B|C) = P(A|B,C)P(B|C)
$$
(you just add a condition on $C$, but otherwise that's the same formula). You can then apply this formula for $A = y$, $B = \theta$, and $C ...
1
Interpretation 1: No red and blue balls within any possible square of 1m x 1m
#### plotting situation
### drawing boxes around the blue points
box <- function(x, y , col = "blue") {
polygon(c(x,x,x,x)+c(-1,-1,1,1), c(y,y,y,y)+c(-1,1,1,-1), border = col)
}
box <- Vectorize(box)
set.seed(1)
### plot the 100 x 100 square
plot(-1000,-1000, ...
0
when $A$ and $B$ are not independent,
$P(B)=P(B|A)P(A)+P(B|\bar A)P(\bar A)=P(A,B)+P(\bar A,B)$.
Thus, mathematically $P(A)P(B)=P(A)[P(A,B)+P(\bar A,B)]$.
If $P(B)\geq P(B|A)$, since $P(B|A)=\frac{P(A,B)}{P(A)}$, then $P(A)P(B) \geq P(A,B)$. Or symmetrically, if $P(A)\geq P(A|B)$, we also have $P(A)P(B) \geq P(A,B)$.
Otherwise, $P(A)P(B) < P(A,B)$.
1
For the empirical rule, we say “two” because it’s more convenient to say that that to say 1.96. The correct value is 1.96, not two. (Even 1.96 has some amount of rounding.)
The 95% for a confidence interval is a separate issue. A 95% confidence interval means that, if you took new samples from your population over and over, if you follow the procedure to ...
2
There are.
If you tell me that "Dice have a 16.7% chance to roll a 1". I can test
that statement to whatever confidence level I like.
Is the classical or frequentist probability, it is related to frequencies of events among all the possible events.
But if you tell me that "There is a 20% chance of rain" or "There is a
45% chance ...
3
Yes, the classical PDF from beginning calculus-based probability is the “Radon-Nikodym derivative” of the probability measure with respect to Lebesgue measure. That more-or-less corresponds to taking the derivative of the CDF to get the PDF.
Even though the intro probability class makes it sound like we get CDFs by integrating densities, CDFs come directly ...
1
In accordance with whuber's suggestion, I am posting an extended version of some comments that I made on whuber's answer as a separate answer of my own.
The experiment consists of players A and B each (independently) tossing their individual coins that turn up Heads with probabilities $p_A$ and $p_B$ respectively. Repeated independent trials of this ...
0
The best Chebyshev inequality (see the reference below) for gamma random variables, where $k \gt E[X]$, is given by $$P \left[ X > k \right] \leq \frac{k f(k)}{k \lambda - \alpha} \ ,$$ with $f(x)$ being the density function $$f(x)=\frac{x^{\alpha-1}\lambda^\alpha e^{-\lambda x}}{\Gamma \left( \alpha \right)}$$
If we reverse your inequality, we are ...
2
just create list of all files, then sample this list and voila!,
as for calculations:
probability sampling image from class $A$ sampling from whole set is $P(class\_A)=size(A)/size(whole\_set)$ but in second sampling/drawing $P(class\_A)=\frac{size(A)-1}{size(whole\_set)-1}$ because I presume you are sampling without replacement, of course after picking ...
1
Good on you for considering effect sizes vs. significancy. The thing is, a p-value tells you, this difference is very surprising if the null-hypothesis (there is no difference) is true. However, it says nothing substantial about if it's actually a big difference. This is a major confusion when you talk about a significant result, because statistically ...
1
The support of $Y$ conditional on the value of $X=x$ is $(-x,x)$ so that should be your region of integration.
0
from markov inequality:
$$P(X \ge a) \le E(X^q)/a^q$$
$$P(X \ge a) = 1 - P(X \lt a) \le E(X^q)/a^q => P(X < a) \ge 1 - E(X^q)/a^q$$
from
https://en.wikipedia.org/wiki/Generalized_gamma_distribution
https://en.wikipedia.org/wiki/Gamma_distribution#General
$$E(X^q) = a \Gamma((d+1)/p)/\Gamma(d/p) \stackrel{d=n/q, a=(1/3)^q, p=1/q}=(1/3)^q\Gamma(n + q)/\...
15
Probability of 'observations' given the 'model'
Typically 'probability' is expressed as the probability of an outcome given a particular experiment/model/setup.
So the probability is about the frequencies of observations given the model. These types of questions are often not so difficult. For instance, in gambling, we can express the probabilities of ...
1
You are interested in the distribution of number of successes, that is how many dice out of an initial $x$ will be left in the game after three rolls, given the rules. You estimated $E[X]=np=20 \cdot 0.11=2.2$, this is the expected number of dies that remain in the game at the end. The variance is $Var[X]=np(1-p)=20 \cdot 0.11(1-0.11)=1.958$.
Your question, ...
1
There are plenty of great answers already, so I'll add a slightly tangential example that I found intriguing. Hopefully its not too far-off from the topic.
Markov chain Monte Carlo methods are often used for Bayesian posterior inference. In typical encounters of Markov chains in probability theory, we ask questions like whether a chain converges to some ...
20
"Inverse probability" is a rather old-fashioned way of referring to Bayesian inference; when it's used nowadays it's usually as a nod to history. De Morgan (1838), An Essay on Probabilities, Ch. 3 "On Inverse Probabilities", explains it nicely:
In the preceding chapter, we have calculated the chances of an event,
knowing the ...
answered Oct 21 at 21:57
2
I'm pretty sure the expected answer to this problem involves an argument such
as the one suggested by @whuber.
However, because the values are specific, a grid search in R can be used to
find an exact solution, as shown below. I assume $3$ is the gamma rate parameter, as in R. [There is no general
agreement whether the second parameter of a gamma random ...
2
$X \sim Bern(0.2)$
By the definition of median
$P(X \leq m) \geq 1/2$ and $P(X \geq m) \geq 1/2$
It has
$m = \begin{cases}
0, \quad p < 1/2\\
[0,1], \quad p = 1/2\\
1, \quad p > 1/2
\end{cases}$
It then follows that $m = 0$.
6
Let $X \sim \mathsf{Bern}(p=.2)\equiv\mathsf{Binom}(n=1, p=.2).$ In R, where qbinom is the inverse CDF (quantile function) of a binomial distribution a median $\eta = 0.$
qbinom(.5, 1, .2)
[1] 0
$P(X \le 0) = P(X = 0) = 0.8 \ge 1/2.$
dbinom(0, 1, .2)
[1] 0.8
And obviously, $P(X \ge 0) = 1 \ge 1/2.$
The CDF of $X$ is plotted below. The median of $X$ is ...
2
From the side of computational complexity
Your sentence:
Someone here said logarithms are too complicated to calculate, but I don't see how hard could it be, given that it is just a button on a calculator. And as said, log probabilities are more stable than multiplied/squared probabilities.
is not accurate in terms of implementation and pragmatism of ...
3
Let $Z$ be a random variable with distribution $P^Z$, meaning that for any measurable set $A$,$$\mathbb P(Z\in A)=P^Z(A)$$
Then, for any measurable transform $f$, $X=f(Z)$ is a random variable with distribution $P^X$ such that, for any measurable set $A$,$$P^X(A)=\mathbb P(X\in A)=\mathbb P(f(Z)\in A)=\mathbb P(Z\in f^{-1}(A))=P^Z(f^{-1}(A))$$
where
$$f^{-1}(...
5
Yes, I believe your thinking is a way to view things in that it points out that the prior is the key ingredient to convert conditional probabilities.
My reading is that it is an interpretation of Bayes' theorem, which, as we know, says
$$
P(B|A)=\frac{P(A|B)P(B)}{P(A)}.
$$
Hence, Bayes' theorem provides the result to convert one conditional probability into ...
2
Perhaps the simplest thing to do here is just to compute the probability of every possible outcome at the end of the three-stage process. For the sake of broader interest, let me generalise your description of the problem by supposing we have a $K$-round process with $n$ initial dice, with $\boldsymbol{\theta} = (\theta_1, \theta_2, ..., \theta_K)$ as the ...
1
First of all, it is a common misconception that a probability distribution only "exists" if it is of one of the forms found in the families of distributions that receive attention in academic literature and are therefore listed on resources like Wikipedia. Every discrete distribution obeying the rules of probability exists, regardless of whether ...
2
For the first question, since $Y_{n}$ is of binomial distribution, the probability that $Y_{n}=y$ is
$P(Y_{n}=y)={n \choose y}\theta ^y\times(1-\theta)^{n-y}$.
The probability that $n-Y_{n}=k$ then is
$P(n-Y_{n}=k)=P(Y_{n}=n-k)={n \choose n-k}\theta ^{n-k}\times(1-\theta)^{k}={n \choose k}(1-\theta) ^{k}\times\theta^{n-k}$.
Thus, $n-Y_{n}$ is of binomial ...
2
For you second question, since you already know that the mean of $Y_{n}$ is $n\theta$, and $n\sim\mathcal{Poisson}(\lambda)$, the mean of $Y_{n}$ is
$E[Y_{n}] = E[n\theta]=\theta E[n]=\lambda\theta$. Or equivalently, $E[Y_{n}]=\sum_{n} n\theta*\frac{\lambda^{n} e^{-\lambda}}{n!}=\lambda\theta$.
Similarly, the various of $Y_{n}$ is $\sum_{n} n\theta (1-\theta)...
1
For actually generating $n$ such values, here are three approaches in R turning uniform pseudo-random variables into your distribution. The first is a simplified inversion of the CDF, putting its two parts together; the second essentially takes an exponential distribution and applies a random sign; and the third is essentially the difference between two ...
2
Your assumption of $y_t \perp \!\!\! \perp \exp(\varepsilon_{t+1})$ is correct by definition of $\varepsilon_t$, i.e:
$X \perp \!\!\! \perp Y \Leftrightarrow X \perp \!\!\! \perp f(Y) \ \forall \ f(\cdot) \tag{1}$
where $\perp \!\!\! \perp$ is the independence symbol. In words: if $X$ is independent of $Y$, then $X$ is also independent of any deterministic ...
2
A simpler way is to evaluate the joint survival at either $x=0$ or $y=0$.
The joint survival function is:
$$
S(x,y)=\mathbb P(X>x,Y>y)
$$
and the marginals survival functions are
$$
S_X(x) = \mathbb P(X>x) \\
S_Y(y) = \mathbb P(Y>y)
$$
since $X$ and $Y$ belong to $(0,1)$ then
\begin{align*}
\mathbb P(X>x) &= \mathbb P(X>x, Y>0) \\
&...
5
This is laplace distribution, you can either use the CDF given in the wikipedia page or find a proof. If you want to integrate it yourself, the integral will be from $-\infty$ (not $0$) to $x$, and for the negative portion, you'll just substitute $|x|=-x$ and do the integration.
1
You have shown no attempted solution so there is no way to know
the exact approach you are intended to take for this assignment.
Many mathematical statistics books have a general formula for
the density function of the $k$th order statistic of a sample
from a distribution in terms of the CDF of the population distribution.
In this case, where the population ...
1
The policy is a function you define to act in an environment, it's not a random variable.
0
Sampe space refer to to $\Omega$ which is the set of all possible elementary events $\omega$. I don't really know what you mean by outcome space. If you mean the space of all possible outcomes, then you are referring to the sample space $\Omega$. If you mean the space, or rather set, of an outcome, you are referring to an event.
In your first example $U$ is ...
4
To understand their relation, you should go back to how $\sigma^2$ is defined. Recall that in the discrete case
$$\sigma^2=Var(X)=E[(X-\mu)^2]$$
If you have have all observations in the population, you can calculate this expected value by the formula you first provided
$$\sigma^2=\frac{1}{N}\sum^N_{i=1}(x_i-\mu)^2.$$
When $X$ instead is a random variable, ...
1
The probability of independent events is the product of their probabilities, so the answer is
$$\underbrace{.3 \times .3 \times \cdots \times .3}_\text {10} = 0.3^{10} \approx 0.000006 = 0.0006\, \%.$$
(The probability that all mice die.)
0
I got a different result from the accepted answer and would like to know where I've gone wrong.
I assumed a fair, 6-sided die, and simulated 1000 runs of 1000 rolls each. When the result of a roll matches the results of the previous 4 rolls, a flag is set to TRUE. The mean of this flag column and the mean of the runs is then reported. I get ~0.07% as the ...
1
Let's assume that the coin is fair, $p_{heads} = p_{tails} = 0.5$. Let's assume that the models are independent in their predictions (*).
Bayesian approach
Then we can make our decisions based on a Bayesian posterior prediction.
The posterior odds ratio for the true coin being heads (H) and the true coin being tails (T) are the likelihood ratio multiplied ...
1
There are many ways to combine the 3 models.
If you have the response variable and predictions for a good number of observations, you can combine them in a way that better predicts the response variable.
One can average them, build a weighted average from a logistic regression, or a tree based approach. Those are called ensembles and are quite common.
I know ...
16
As I'm sure you know (or can easily derive), the maximum likelihood estimator of $(\mu,\sigma^2)$ is
$$(\hat\mu,\hat\sigma^2) = \left(\bar X, \operatorname{Var}(X)\right)$$
where, as usual, $n\bar X = X_1+X_2+\cdots + X_n$ and, a little unusually,
$$\operatorname{Var}(X) = \frac{1}{n}\left((X_1-\bar X)^2 + (X_2-\bar X)^2 + \cdots + (X_n-\bar X)^2\right)$$
(...
6
My question is, how does this reconcile with the fact that the probability of having 6 girls out of 6 births from the binomial thereom is far less likely than 3 girls (0.015625 vs 0.3125)?
The probability of 3 girls is the probability of bbbggg, bbgbgg, bbggbg, bbgggb, bgbbgg, bgbgbg, bgbggb,gbbbgg, gbbgbg, gbbggb, bgggbb, bggbgb, bggbbg, gbggbb, gbgbgb, ...
1
By independence and standard rules for variances,
$$
Var(\bar{X})=\frac{1}{n^2}Var\left(\sum_iX_i\right)=\frac{1}{n^2}\sum_iVar(X_i)
$$
Given identical distributions,
$$
\frac{1}{n^2}\sum_iVar(X_i)=\frac{n}{n^2}Var(X_1)\to0
$$
Hence, the variability in the sample mean disappears, so to speak, when computing it over many realizations of the underlying random ...
0
A random variable $X$ is defined in terms of possible outcomes $\Omega$, not observed outcomes from trials. If $\Omega$ is defined to be countable, then $X$ is discrete, and if $\Omega$ is defined to be uncountable, then $X$ is continuous. Any finite set of draws of $X$ will not only be countable but finite, whether $\Omega$ is uncountably infinite, ...
0
Height is continuous in principle, but reported as one of various discrete measurements in practice.
What's more, there can be many minutely different conventions both within and between datasets.
For example, in some countries original measurements might be variously in inches or cm (mm) and standardized to one or the other. That can lead to a distribution ...
1
Monte Carlo method
DyedPurple already showed that your simulation is not wrong and you should get a probability of ~0.84 for a run length of 1000. It is only when the run length goes towards infinity that you are almost certain to get gambler's ruin (If you have a stopping rule for some upper boundary, as in this question, then you can escape the gambler's ...
4
.. each with a specific height that can be measured with infinite
accuracy..
Based on this, we could say that height of a single individual, say $X$, is continuous RV, it can be any real number within a plausible range. This makes vector of heights, say $X^N$, a continuous random vector as well.
For example, we could have 𝑁=3 people with heights 150 cm, ...
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