Find conditional probability of $E[X_{1}|X_1+X_2+...+X_{n}]$ given the   fact that $X_{k}$ for all $k$ have finite expected value and all the   variables are independent and have the same distribution.   I guess the solution should follow quickly from one of the properties of the conditional probability. But I don't see any that would be helpful. So is applying the definition the right way?
24993
I don't really know how to start proving this question.  Let $\xi$ and $\eta$ be independent, identically distributed random variables with $E(|\xi|)$ finite.  Show that $E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)=\frac{\xi+\eta}{2}$  Does anyone here have any idea for starting this question?
43100
I'm having a little trouble with the proof that the expected value of $x_i$ is $ \bar{X} $.  What I have is   $E[x_i]=\sum_{j=1}^{N}X_j Pr(x_i=X_j) $  Then,  $Pr(x_i=X_j) = 1/N $  This is the bit I can't understand, how does that probability evaluate to that value.  I know the denominator is how many ways you can choose n out N. I think that the numerator should be how many ways you can choose (n-1) out of (N-1). But I seem to have an extra n.  Any advice ?
103005
I am looking for conditional expectation $E[X| X&gt;Y]$ for independent $X, Y \sim N(0,1).$ I just showed that $P(X\leq a | X&gt;Y) = \Phi(a)^2$, where $\Phi(a)$ is CDF for Normal$(0,1)$, but I don't know how to compute integral in order to obtain expected value.
110506
Is there any counterexample to show that $X,Y$ are two random variables and $E(X\mid Y)=E(X)$, but $X$ and $Y$ are not independent.   I already know that if $X,Y$ are independent, then $E(X\mid Y)=E(X)$.
41737
E, F and G are independent events with probabilities $\frac12,\frac14,\frac15$   of occurring. Let $X$ be the number of events that occur.      A. Describe $X$ and calculate its expectation and variance.   B. The same, but this time assume the events are mutually exclusive.   I really don't know how to proceed. I have tried to set the support of $X$ as $\{0,1,2,3\}$ and then write the probability of each event as $P[E](X=1)=\frac12$ but I don't think it's the real solution.
98938
Let $X$ be a non-negative integer-valued random variable with finite mean. Show that $$E(X)=\sum^\infty_{n=0}P(X&gt;n)$$  This is the hint from my lecturer.  "Start with the definition $E(X)=\sum^\infty_{x=1}xP(X=x)$. Rewrite the series as double sum."  For my opinion. I think the double sum have the form of $\sum\sum f(x)$, but how to get this form? And how to continue?
59056
Setting  Suppose X1 and X2 are independent with the common geometric distribution w(k; p). Determine the conditional distribution of X1 given that X1 + X2 = n.  Solution  My argument is   $$\Pr[X_1| X_1+X_2 = n] = \frac{\Pr[X_1 + X_2 = n| X_1 = x_1]\Pr[X_1 = x_1]}{\Pr[X_1+X_2=n]}$$  and  $$\Pr[X_1 = x_1] = q^{x_1} p$$  $$\Pr[X_1+X_2=n|X_1=x_1] = \Pr[X_2 = n - x_1] = q^{n-x_1}p$$ while $$\Pr[X_1+X_2=n] =\sum_{j=1}^n q^j p q^{n-j} p = q^n p^2$$  However, I am not confident with this solution since geometric variables should be memoryless? So I feel like I am missing something here.
53110
Could anybody help me with solving such problem:  You are given two independent random variables $X$ and $Y$ with continuous uniform distribution. You are to find expected value $E$ of random variable $Z = \max(X,Y)$.
74362
If $x_1,x_2,x_3$ and $x_4$ are independent identically distributed  random variable with continuous distribution .what is the probability of $$P=p(x_1 &lt; x_2&lt; \max(x_3,x_4))$$ I know the exhaustive number of cases are $p(x_1&lt;x_2&lt;x_3)+p(x_1&lt;x_2&lt;x_4)$ and  every $x$ are iid so both have same probabiltiy so is $2p(x_1&lt;x_2&lt;x_3)$ is right? But  i get the $p(x_1&lt;x_2&lt;x_3) = 1/6$ and then my answer is $1/3$ but the answer is $1/6.$ Can we write $P$ as $p(x_1&lt;x_2&lt;x_3&lt;x_4)+p(x_1&lt;x_2&lt;x_4&lt;x_3)$?
76920
Suppose for two random variables $X_1$ and $X_2$ $\sigma(X_1) = \sigma(X_2)$. Why $E[Y | X_1] = E[Y | X_2]$ a.e. ?  the set where $X_1= X_1(\omega)$ can be different from the set $X_2= X_2(\omega) $. How can then conditional expectation conditioned on the event can be equal ?
45975
The problem:  X is a binomial random variable, find $E[\frac{1}{X+1}]$ n and p are not given  PDF for a binomial distribution is $\binom{n}{k}p^k(1-p)^{n-k}$  Expected value is $\sum{x_ip(x_i)}$  But this is where I get stuck, I'm really rusty on my statistics and I'm not sure exactly how to structure it in the next step? I think I want to get the form of the following out of the summation  $\sum _{k=0}^{n} \binom{n}{k}p^k(1-p)^{n-k} = (p + 1 - p)^n = 1$  But I'm not sure if it should look like  $\sum \frac{1}{xp(x)+1} $  and if it should where to go from here?
38822
The questions asks to $ E[X_1|S_n]$ where $ S_n = \sum_{[n]} X_i $ with $X_i$ i.i.d. of finite expectation. My attempt was to consider an arbitrary Borel set, pull it back under $ S_n $ to get a set in $\sigma(S_n)$, and then average $X_1$ over that pull back and normalize:  $$ E[X_1|S_n] = E[X_1 1_{S_n^{-1}(B)}]/P[S^{-1}_n(B)]$$  for any Borel set $B$.  However, it just doesn't feel right - it seems just symbol pushing, without giving any intuition to what's happening.   Can anyone explain if this is correct, of if I missing something deep?
111581
Suppose $X,Y$ and $Z$ are multivariate normal with means and full covariance matrix. The conditional expectation $E(X | Y)$ is well know. What is the conditional expectation of $E(X | Y,Z)$ if $Y$ and $Z$ (and $X$) are correlated? Standard textbooks only seem to cover the case when $Y$ and $Z$ are uncorrelated.
53948
Doing a little reading over the break (The Probabilistic Method by Alon and Spencer); can't come up with the solution for this seemingly simple (and perhaps even a little surprising?) result:     (A-S 1.6.3) Prove that for every two independent identically distributed real random variables $X$ and $Y$,   $$Pr[|X-Y| \leq 2] \leq 3 Pr[|X-Y| \leq 1].$$
29367
When $X$, $Y$ are bivariate normal, how can we calculate the conditional expectation $E(X\mid X&gt;Y)$? A more concrete setting of this general problem can be referred .
24230
$X$ and $Y$ are independent $U(0,1)$ random variables.   I have to find $E[X|X&gt;Y]$.   I found it out by integrating $\int_0^1 \int_y^1 x dxdy$ to get $\frac{1}{3}$ as the answer but this is wrong according to the answer key. What is the mistake that I have made?
119958
An exercise asks me to show that if $X_1,X_2,\ldots,X_n$ is a random sample from a Poisson distribution with parameter $\theta$, the conditional distribution of $X_1,X_2,\ldots, X_{n-1}$, given $Y=\sum_{i=1}^n X_i$, is multinomial. Normally, this would have been an easy one but unfortunately the joint distribution here throws me off a bit. This is where I would appreciate some help.