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5 votes

How do I compute a probability from the MGF?

One approach is to say that if the moment generating function is $$\mathbb{E}\left[e^{tX}\right] = \frac{2}{9} + \frac{e^{-t}}{9} + \frac{e^{-2t}}{9} + \frac{2e^{t}}{9} + \frac{e^{2t}}{3}$$ then, ...
Henry's user avatar
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3 votes
Accepted

How do I compute a probability from the MGF?

The MGF of a random variable $X$ is defined as $m_X(t) = \mathbb{E}[e^{tX}]$, where $\mathbb{E}$ denotes the expected value. The MGF can be used to determine all the moments of the distribution (like ...
ADAM's user avatar
  • 577
6 votes

How do I compute a probability from the MGF?

Hint: For a discrete random variable with values in $\mathscr{X}$ the moment generating function is: $$m_X(t) \equiv \mathbb{E}(e^{tX}) = \sum_{x \in \mathscr{X}} e^{tx} \cdot \mathbb{P}(X=x).$$ Does ...
Ben's user avatar
  • 123k
2 votes

Is there a closed form CDF for the sum of two triangularly distributed random variables?

Following on @WindFish's answer, it is possible to derive the closed form solution, but it is very detailed. For each $z$ you need to keep track of the regions of $x$ and $y$ to determine which part ...
R Carnell's user avatar
  • 5,093
0 votes

Two random variables X1 and X2 may be partially dependent i.e. X1 is independent of X2 but X2 is dependent on X1?

Two random variables X1 and X2 may be partially dependent i.e. X1 is independent of X2 but X2 is dependent on X1? No. Statistical dependence/independence is always a two-way relation, because it is ...
Alecos Papadopoulos's user avatar
2 votes
Accepted

Number of simulated statistics more extreme than most extreme real data statistic

Depending on the details (if realisations are without error) then you can simplify this question as drawing two samples of size $n$ and $m= qn$, and express the distribution for the number of times ...
Sextus Empiricus's user avatar
2 votes

A and B are independent. Does P(A ∩ B|C) = P(A|C) · P(B|C) hold?

A related question is If $X, Y$ are independent of $Z$, is $P(X|Y, Z) = P(X|Y)$? The example there, $C = XOR(A,B)$, is a simple counter example to this question as well ...
Sextus Empiricus's user avatar
1 vote
Accepted

Optimization of fault diagnosis sequence using probability and cost

The optimal order for the tests is the same as the order of the cost to price ratios. Consider just two tests, $T_1$ and $T_2$ with probabilities $p_1$, $p_2$ and costs $c_1$, $c_2$. The sequence $\{...
jblood94's user avatar
  • 1,436
1 vote
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Understanding three prisoners (Statistical Inference - Cassella and Berger)

The wikipedia page for the problem helps pinpoint the reason for the 1/6 probability which I quote below. It also articulates the problem setting in a bit of a more verbose way which helps understand ...
giorgio's user avatar
  • 126
0 votes

Variance of Multimodal Generalized von Mises Distribution?

While I suspect that there is no nice closed-form for any of the moments (and also the constant of integration), until someone comes up with a good solution having a numerical approximation as a check ...
JimB's user avatar
  • 3,664
2 votes

Probability to get a 1 in 5 rolls 3 or more times

Your math is not right. You are missing part of the binomial probability formula. The correct calculation would be $$P(X = 3) = \binom{5}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^2 \approx ...
wzbillings's user avatar
22 votes

A and B are independent. Does P(A ∩ B|C) = P(A|C) · P(B|C) hold?

No this is not in general true, as you can see from a simple counter example: Toss two independent coins. Event $A$ is coin 1 head. $P(A)=0.5$ Event $B$ is coin 2 head. $P(B)=0.5$ Event $C$ is either ...
George Savva's user avatar
  • 2,034
-3 votes

What is so cool about de Finetti's representation theorem?

I had a student who asked me: " isn't de Finetti Theorem an asymptotic result? Who cares ? " ( as you know, young Bayesians tend to be more bayesian than Bayes himself. Lol) This made me ...
Quarentinha's user avatar
1 vote
Accepted

Calculating $E[(\sum X_i)^4]$

In pursuit of calculating the fourth power, it might be apt to have a look at the general form of expectation of $p$–th power of sum of iid random variables. $\rm [I]$ deduces $\mathbf E\left[S_n^p\...
User1865345's user avatar
  • 7,847
3 votes

Calculating $E[(\sum X_i)^4]$

I would start by setting $Y_i=X_i-\mu$, for $i=1\ldots,n$. You then have: $$ \mathrm{Var}(\bar{X}^2)=\mathrm{Var}((\bar{Y}+\mu)^2)= \mathrm{Var}(\bar{Y}^2+2\mu\bar{Y})\\ = \mathrm{Var}(\bar{Y}^2)+2\mu\...
Doctor Milt's user avatar
  • 2,672
2 votes

Can we conclude from $\DeclareMathOperator{\E}{\mathbb{E}}\E g(X)h(Y)=\E g(X) \E h(Y)$ that $X,Y$ are independent?

This is almost Proposition 7.1.3(i) from Athreya & Lahiri 2006 sans some minor differences in formatting: Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $\{X_1, \ldots, X_k \}$, $2 ...
Galen's user avatar
  • 7,985
0 votes

The probability of a random variable being larger than a sequence of random values

As mentioned by others you should notate it with capitals $X_{1}, X_{2}, \dots, X_{n}$. Let $A_{i}$ be the event $X_{i}$ is the largest. Then we have $1 = P(A_{1}) + P(A_{2}) + \dots + P(A_{n})$. ...
Elmex80s's user avatar
  • 101
1 vote

Using R for dice probabilities

These are arithmetically very simple problems. Therefore, it's not clear in what way you intend to use R as a calculator to solve your problems for you. Focusing on example 3 as the hardest example: ...
AdamO's user avatar
  • 61.9k
0 votes

Using R for dice probabilities

...
Scarlett's user avatar
3 votes

If $X \sim\textrm{ Bin}(100, 0.5), $ then what is the approximate distribution of $(X/5 - 10)^2? $

$X \sim \text{Binomial}(100, 0.5)$ can be approximated by a Normal distribution with a mean of $100 \times 0.5 = 50$ and variance of $100 \times (0.5) \times (1 - 0.5) = 25$. Therefore $\frac{X}{5}$ ...
Eoin's user avatar
  • 8,812
2 votes
Accepted

Why is pnorm giving different results with X, sd and mean vs corresponding z value?

The inconsistency stems from the use of $\sigma=5$ in the pnorm call, whereas your original specification and second formula have $\sigma^2=5$. The usual notation ...
PBulls's user avatar
  • 3,558
2 votes
Accepted

Can we write $\int_p^\infty\int_q^\infty f(x,y)dxdy=\int_p^\infty\int_q^\infty\frac{\partial^2}{\partial x\partial y}F(x,y)=F(\infty,\infty)-F(p,q)$?

A geometrical interpretation will show that, because $F(p,q)=P(X\lt p \cap Y \lt q)$ $$P(X\gt p \cap Y \gt q)=1-P(X<p)-P(Y<q)+P(X\lt p \cap Y \lt q)\\ =1-F_{X,Y}(p,+\infty)-F_{X,Y}(+\infty,q)+F_{...
Firebug's user avatar
  • 18.9k
4 votes

How to incorporate prior knowledge after ML training?

You can either use an ensemble of multiple classifiers that will improve performance when your ML classifier breaks down[1], or implement boosting for your ML classifier assuming that it's a "...
Leif Peterson's user avatar
3 votes

Can a ML classifier's prediction be understood as a probability?

That would be desirable, but it is not guaranteed to make as much sense as we might like. First, you could make an argument that any predicted $p(\mathcal C_k|\mathbf x_i)\in[0,1]$ is a probability in ...
Dave's user avatar
  • 60.7k
3 votes

How many A is needed to be B number of results with a C probability?

Take a look at the Negative Binomial distribution. The probability the action succeeds is 15%, and you are repeating this action until you observe 5 successes. This corresponds to parameter values $p=...
Doctor Milt's user avatar
  • 2,672
1 vote

Deriving the Distribution of Markov Chain Times

Would love to also understand how closed-form solutions for the variances of these distributions can also be calculated (i.e. non simulations) if possible. Below is an example of Markov chain with 5 ...
Sextus Empiricus's user avatar
5 votes

Deriving the Distribution of Markov Chain Times

I can show you how to derive the discrete-space result, and the extension to the continuous-space result is essentially a limiting version of this. Distribution of absorption time: The formula you ...
Ben's user avatar
  • 123k
1 vote

Infinite dice roll probability

Another graphical way to go about this is to use a familiar tree, We see that the probability of $A$ winnning per game is $6/72$ and the probability of the game repeating is $50/72 + 5/72 = 55/72$. ...
Alecos Papadopoulos's user avatar
0 votes

Infinite dice roll probability

I am sorry because I do not know Markow chains and such theorems because I am kind of new to high level math, but this probability question is something that I would like to answer and maybe, the ...
Ham Lemon's user avatar
  • 101
1 vote

Infinite dice roll probability

We don't need matrices — this can be modeled with series. Let $P_W$ denote the probability $A$ wins on their roll, and $P_L$ denote the probability $A$ loses but gets another chance to roll (i.e., $B$ ...
Stanley Yu's user avatar
0 votes

A question on distributivity of independence: $A\perp(Y,R) \implies A \perp Y $ and $A \perp R$?

Let ${\cal A}$ be a set that $A$ could be in, ${\cal Y}$ a set that $Y$ could be in, and ${\cal R}$ the entire sample space for $R$ Then $$P(A\in {\cal A} \cap Y \in{\cal Y})= P(A\in {\cal A} \cap Y \...
Thomas Lumley's user avatar
1 vote

Infinite dice roll probability

@Glen_b and @whuber have gave good answers, and I think the gist is the Bayesian formula. Here is another method: For a single round , the probabilities are: ...
lishuang wang's user avatar
0 votes

Why do we work with factor of likelihoods instead of e.g. a sum for a batch in the negative log likelihood loss function?

The "likelihood" is a derivative concept. The primary concept is probabilistic: we consider the joint density of a sample of independent observations, because it expresses the probability of ...
Alecos Papadopoulos's user avatar
17 votes

Infinite dice roll probability

Evidently $A$ has a $3/6 \times 1/6 = 1/12 = a$ chance of winning on their turn and $B$ has a $1/6 = b$ chance of winning. From $A$'s perspective, their chance $p_A$ of winning this game is the sum of ...
whuber's user avatar
  • 321k
12 votes
Accepted

Infinite dice roll probability

The matrix doesn't seem right, but it is difficult to see as the labels are not very clear. What is the transition from 1 to ...
Sextus Empiricus's user avatar
3 votes

Why do we work with factor of likelihoods instead of e.g. a sum for a batch in the negative log likelihood loss function?

Commonly, in statistics, we make a strong distributional assumptions about where each datapoint comes from: we say that $P(y_i|\theta) = f(y_i,\theta)$. If we make the assumption that the data are ...
John Madden's user avatar
  • 4,055
2 votes

Advantages of using Vine Copulas over Regular Copulas?

Copula: Very complicated in multivariate cases. Not all copula can be extended into multivariate copula. It imposes the same dependency among all variables, not the case for many real-life data. ...
0 votes

Why do we work with factor of likelihoods instead of e.g. a sum for a batch in the negative log likelihood loss function?

This relates to the questions: Could a mismatch between loss functions used for fitting vs. tuning parameter selection be justified? and If the predicted value of machine learning method is E(y | x), ...
Sextus Empiricus's user avatar
0 votes

Name of PDF? - projecting uniform probability distribution on the unit circle to the x-axis

If $X$ and $Y$ are two independent standard Gaussian distributions, then the random point $\left(\frac{X}{\sqrt{X^2+Y^2}}, \frac{Y}{\sqrt{X^2+Y^2}}\right)$ has the uniform distribution on the unit ...
Stéphane Laurent's user avatar
0 votes

Conditional expectation in the multivariate normal distribution

Let's start with the properties of the multivariate Normal distribution and see where that leads us. The "unconstrained" conditional expectation is a random variable, say $W$, and a function ...
Alecos Papadopoulos's user avatar
1 vote
Accepted

what is actually the difference between prior and evidence?

Bayes theorem says: $P(A|B) = P(A) \times \frac{P(B|A)}{P(B)} = \frac{P(A, B)}{P(B)}$. In this, A is as you say your 'prior belief' and B is your 'evidence'. Bayes theorem thus updates your prior ...
Mathemagician777's user avatar
6 votes
Accepted

Is there a mathematical function with a shape like an inverted sigmoid pdf but X values that go beyond 1?

A one-parameter family of functions that seems to do what you want is one minus the quantile function of the beta distribution, up to rescaling $x$ by dividing by 150. "One-parameter", ...
Stephan Kolassa's user avatar
0 votes

How can model overconfidence coincide with accurate classifications?

Just because a model is highly accurate, it doesn't guarantee a high level of confidence. Take for example the predicted class probabilities for the e.g. $i$th object in a 3-class problem from the ...
Leif Peterson's user avatar
2 votes

Prove that the equality holds

If you know the analogous formula for variances, you can apply it to the identity $$4\mathrm{Cov}[X,Y]=\mathrm{var}[X+Y]-\mathrm{var}[X-Y]$$
Thomas Lumley's user avatar
7 votes

The Bing Tibetan Glitch Emoji Problem

The poisoned wine problem --- combinatorial group testing The problem stated here is not really a statistical problem --- it is primarily a combinatorial problem that is sometimes referred to as ...
Ben's user avatar
  • 123k
3 votes

How to test the statistical significance of probabilities?

I think you might have a bit of a misconception about what statistical significance actually means. Statistical significance only makes sense if you have data from a sample/samples where you don't ...
LevG's user avatar
  • 173
2 votes
Accepted

Do "Likelihood Properties" apply to "Non-Likelihood Solutions"?

Indeed for some maximum likelihood estimators we have $$\sqrt{n}(\hat\theta - \theta) \to \mathcal{N}(0,I^{-1})$$ from which those properties follow. What I am getting at, is that inevitably, ...
Sextus Empiricus's user avatar
1 vote

Theory of 'Runs' and probability

Is this answer computed by me and its method of computation correct? No, because it doesn't have a sample space of equally-probable outcomes. The stated problem is to find the probability that there ...
r.e.s.'s user avatar
  • 301
3 votes

The Bing Tibetan Glitch Emoji Problem

Here is my partial answer thus far. This kind of uses information-theoretic/Bayesian reasoning but I think you could probably reformulate it in frequentist terms. One can think about, for instance, ...
Mike Battaglia's user avatar

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