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Hot answers tagged conditional-probability

9

The transition from the second to the third line does not follow. Since $\mathbb{E}(X) \neq \mathbb{E}(X|Y)$ you have: $$\mathbb{E}[ (X - \mathbb{E}(X))^2 | Y ] \neq \mathbb{E}[ (X - \mathbb{E}(X|Y))^2 | Y ] = \mathbb{E}[ \mathbb{V}(X|Y) ].$$ In the special case where $\mathbb{E}(X) = \mathbb{E}(X|Y=y)$ for all $y \in \mathbb{R}$ your working and ...

6

The third line is wrong, because you don't have $\text{E}[X|Y]$ in the second line. For example, if $Y$ is Bernoulli(1/2) and $X$ is 1 if $Y$ is 1 and -1 if $Y$ is 0, then $\text{E}[(X-\text{E}[X|Y])^2|Y] = 0$ (this is what you want) because $Y$ is totally informative of $X$, but what you have will give you $\text{E}[(X-\text{E}[X])^2|Y] = \text{E}[(X-0)^2|Y]... 4$P(\text{white ball}|\text{get 6}) + P(\text{white ball}|\text{not 6})just summates the probability of obtaining a white ball in the scenario you obtained six with the probability of obtaining a white ball in the scenario you did not obtain six. You can see that this expression is incorrect by two ways. First, it is not a valid probability. For example, ... 3 \require{cancel} \begin{aligned} \operatorname{Var}(X) &= \operatorname{E}(X - \operatorname{E}X)^2 \\ &= \operatorname{E}\left(\operatorname{E}\left[(X - \operatorname{E}X)^2\mid Y\right]\right) \\ &\ne \operatorname E\left( \operatorname E\left[ (X-\operatorname E(X\mid Y))^2 \right] \mid Y \right) \\ &= \operatorname{E}(\operatorname{Var}... 2 You can do this with no calculation. As explained (without calculation) at https://stats.stackexchange.com/a/261926/919, the union of two Poisson processes of rates \lambda and \mu is a Poisson process of rate \lambda+\mu: just remove the colors from the points in the top picture and treat them all as coming from the same process, as shown in the ... 2 It is unlikely to be true As a simple example, if X,Y,Z are all mutually independent then P(X\mid Y\cup Z) = P(X) = \frac12\big(P(X\mid Y) + P(X \mid Z)\big) If you really want P(X\mid Y\cup Z) then note it is \frac{P(X\cap (Y\cup Z))}{P(Y\cup Z)}. For the denominator, if Y and Z are independent then P(Y\cup Z) = P(Y)+P(Z)-P(Y)P(Z), ... 2 The solution is: P(Y\in [a,b]|X\in[a,b])=\frac{\int_{[a,b]}(F_{Y|X}(b|x)-F_{Y|X}(a|x))f_X(x)dx}{\int_{[a,b]}f_X(x)dx} This can be seen, as: P(Y\in[a,b]\cap X\in[a,b])= P(Y\leq b\cap X\in[a,b])-P(Y\leq a\cap X\in[a,b])= \int_{a}^b \left( \int_{-\infty}^bf_{Y|X}(y|x)dy \right)dx-\int_{a}^a \left( \int_{-\infty}^af_{Y|X}(y|x)dy \right)dx= \int_a^bF_{Y|X}(... 1 No. Consider P(X) = 1 regardless of Y or Z. Then P(X|Y) + P(X|Z) = 2, which does not equal P(X|Y\cup Z)=1. 1 The first identity of the joint probability with the product of marginalsP(X,Y) = P(X) P(Y)$$only holds when X and Y are independent. Based on the table and the fact that the product of the marginals do not equal the joint probabilities, should simply lead you to suspect a lack of independence. The second identity is true by the definition of ... 1 Intuitively, if V is known, the only random component in X will be W and, in Y, it will be Z. Since the two are independent, so are X and Y. More mechanically, you can start from your and @Jesper's comments:$$\begin{align}P(W=w,Z=z|V=v)&=\frac{P(W=w,Z=z,V=v)}{P(V=v)}\\&=\frac{P(W=w)P(Z=z)P(V=v)}{P(V=v)}\\&=P(W=w)P(Z=z)\end{align}$$1 To be specific, that is MGF of binomial distribution. Let X\sim B(n,p) be a binomial RV, which can be considered as sum of independent Bernoulli trials, X=X_1+...X_n, all iid with \text{Ber}(p):$$E[e^{tX}]=E[e^{t(X_1+...+X_n)}]=E\left[\prod_{i=1}^n e^{tX_i}\right]=\prod_{i=1}^n E[e^{tX_i}]=E[e^{tX_1}]^n$$Bernoulli MGF is easy:$$M_{X_1}(t)=E[e^{tX_1}... 1 If the distribution of something you observe does not depend on a parameter, it cannot possibly give you information about it. Now, if the distribution ofX$depends on the parameter$\theta$and the distribution of$X$given the sufficient statistic$S$does not, it must be the case that all information about$\theta$is in$S$; once the value of$S$is ... 1 Intuitively there is no variance of a known quantity because it doesn’t vary; and if we condition on a RV, it means we’re given the RV, i.e. we know its value. If someone gives it to me, there will be no uncertainty left associated with it. More mechanically: $$var(X|X)=\underbrace{E[X^2|X]}_{X^2}-(\underbrace{E[X|X]}_X)^2=X^2-(X)^2=0$$ 1 The expectation is wrong because$11/6$is too close to$2$. The PDF has a trapezoidal shape, increasing as$y$increase; therefore the mean could have been$2/3$at max. Specifically, it'll be$11/9$: $$E[Y|X=1]=\int_0^2y(1+2y)/6 dy=1/6(y^2/2+2y^3/3)|_0^2=11/9$$ The rest can be solved by your way, but I think using$E[Y^2|X=1]$will be a bit simpler to do. 1 You sum, or integrate, over the joint distribution, in this case of whether the your dog is hungry or not, and whether you feed your dog, or not. Since$P(A, B) = P(A \vert B) P(B)$, you find that $$P(H) = P(H \vert F)P(F) + P(H\vert NF)P(NF) = \frac{1}{10}\frac{6}{10} + \frac{7}{10}\frac{4}{10} = \frac{17}{50}.$$ By total probability we also have ... 1 The second approach is correct. In the first approach, the following conclusion is too direct: It's simply the expected value of$p_{𝐴|𝐵}$We're saying that$P(A|B)=E[p_{A|B}]$, but at first,$p_{A|B}$needs a proper definition, which in turn should, maybe, demystify other terms:$f_{A|B}(p_A,p_B)$. The second approach follows directly from total ... 1 You have no information to generalize to all basketball players. If all you know is the height of Japanese people and the height of Japanese basketball players, you could at best find out the difference between Japanese basketball players and Japanese people in general. To find out the distribution of height of all basketball players, you would first ... 1 Let's say for simplicity that each variable$A$,$B$and$C$has only two states such that they jointly have$8$states. As described in the answers to What is the number of parameters needed for a joint probability distribution? you will need$8-1=7$independent parameters to describe this distribution. You only have$5$parameters$P(A)$,$P(B)$,$P(C)$, ... 1 This is actually asking if we can find the joint with marginals and$P(A|C),P(A|B)$, i.e. knowing$P(C|A)$and$P(B|A)$doesn't provide additional information since they can be found via Bayes rule. First of all, we know nothing about the dependence relation between$B$and$C$, and also we don't know what happens when we condition on two of the events, ... 1 The combinatorics of this problem look particularly simple, so let's see how much we can progress by tackling it head-on: There are$\binom{nr}{r,r,\ldots,r} = \frac{(nr)!}{(r!)^n}$equally likely ways to partition$nr$objects into groups of size$r,r,\ldots,r.$There are$n! = \binom{n}{1,1,\ldots,1}$ways to partition the first$n\$ objects, one per group....

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