I am looking for the mean and variance of a bernoulli distribution of two normally distributed random variables.  An example would be as follows: A coin flip   heads: draw from a $N(\mu1,\sigma1)$ tails: draw from a $N(\mu2,\sigma2)$   For the expected value of the coinflip I would think it equals $p*\mu1+(1-p)*\mu2$. Where $p$ is 0.5 in this case.  But I have no idea how to compute the variance of this bernoulli distribution.  Could somebody help me with this?
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Say I have two normal distributions A and B with means $\mu_A$ and $\mu_B$ and variances $\sigma_A$ and $\sigma_B$. I want to take a weighted mixture of these two distributions using weights $p$ and $q$ where $0\le p \le 1$ and $q = 1-p$. I know that the mean of this mixture would be $\mu_{AB} = (p\times\mu_A) + (q\times\mu_B)$.  What would the variance be?    A concrete example would be if I knew the parameters for the distribution of male and female height. If I had a room of people that was 60% male, I could produce the expected mean height for the whole room, but what about the variance?
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Let $X_1, X_2,\ldots, X_n$ be iid bernoulli r.v. with parameter $p$. Let $S=X_1+\cdots+X_n$ and $Y=X_1X_2$. Compute $\mathbb{E}(Y\mid S)$.  I know that $\mathbb{E}(X_1\mid S) = S/n$. So If I could split $\mathbb{E}(Y\mid S)$ into two then I would be done. But I don't think that is allowed. How to proceed?
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Let $X \sim Bin(n,p)$   Part A.  Show that the argument      "Then $X/n \sim Bin()$ with $E[X/n]=p$and $Var[X/n]=pq/n$"    is false.   My book says we can prove this result using a moment generating function. But to me it makes no intuitive sense: We have Bin-distributed RV $X$ and if we rescale it with $1/n$ of course it will be Bin-distributed.  And we know from standard formulas that $E[aX]=aE[X]$ and $Var[aX] = a^2Var[X]$ so I think the argument is correct.   (For full disclosure, the argument we should falsify according to the book is "Let X be a binomial random variable with mean np and variance np(1−p). Show that the ratio X/n also has a binomial distribution with mean p and variance p(1−p)/n." But I think the argument is correct. )    Ok, now I get part A. Here is Part B. Prove it using the moment generating function of $X/n$.   Attempt: I get $$m_X(t)=E[\exp(tX/n)]=(m_X(t))^{1/n}$$
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I have four random variables, A, B, C, D, with known mean and variance. As well:   Cov(A,B) is known and non-zero  Cov(C,D) is known and non-zero  A and C are independent   A and D are independent  B and C are independent   B and D are independent   I then create two new random variables:   X = A*C Y = B*D   Is there any way to determine Cov(X,Y) or Var(X+Y)?   If not exactly, is there any way to estimate it? What if I could determine the distributions of A and B and C and D?
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Let $Y_1$ and $Y_2$ be independent random variables with   $Y_1\sim N(1,3)$ and $Y_2 \sim N(2,5).$      If $W_1=Y_1+2Y_2$ and $W_2=4Y_1-Y_2$   what is the joint distribution of $W_1$ and $W_2$?   Is my procedure correct?  $E(W_1)=E(Y_1+2Y_2)=E(Y_1)+2E(Y_2)=1+2\cdot2=5$  $Var(W_1)=Var(Y_1+2Y_2)=Var(Y_1)+4 Var(Y_2)=3+4\cdot 5=23$  $f_{W_1}(w_1)=\frac{1}{\sqrt{2\pi}23}\epsilon^{\frac{-1}{2}(\frac{x-5}{23})^2}$  I can't conclude.
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The following question is on my homework assignment that I cannot figure out:     Let U and V be independent random variables, each having a normal distribution with mean zero and variance one. Find the moment generating function of the random variable W = UV .   I have looked around online, and cannot find an answer to this question. In fact, the only answers I can find that even relate to the product of standard normal random variables are using techniques that we never covered in my class. We covered in class how to find the MGF for linear combinations of random variables, by W isn't linear, its a product of two normals. So that technique won't work.   What am I supposed to do? I am completely at a loss. I have tried multiplying the MGFs of U and V together, but that leaves me with something ugly that I can't reduce.
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I have a distribution $X$. By playing around with random samples from $X$, I've determined that $Var(X^i) &gt; Var(X)$ where $i &gt; 1$. However, I can't seem to find a formula for the expected variance of $X^i$, or why it should be greater than $Var(X)$.  Moving away from normal distributions, should it generalize that the scale parameter of any $X^i$ will be greater than $X$?
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I have $k$ pairs of samples from different distributions. For each pair of samples I want to check if the samples are taken from a normal distribution with the same mean and variance. I assume the normal distributions for each pair is different and independent of other pairs. I do not know the mean and variance of any of the distributions.  I am doing a two sample t-test for each of the pairs. I want to combine the $k$ resulting p-values.   I could use Fisher's method but I am wondering about the distribution of the average t-value. If these were z-values rather than t-values I would use Stouffer's z-test. I could also approximate the t-distribution with a z-distribution and than use Stouffer's z-test, but I am not sure if that is a good approximation.  Is there a test like Stouffer's z-test for t-values?  Is there a better way to combine the t-values?
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I've been given a problem where I have  $$ X \sim \mathcal{N}(\mu = 2, \sigma^2 = 9) $$  $$ Y \sim \mathcal{N}(\mu = 3, \sigma^2 = 4) $$  Their correlation is $ \rho_{XY} = 0.6 $. First I am asked for the mean and standard distribution of X+Y, which I found as  $$ \mu_{X+Y} = \mu_X + \mu_Y = 2 + 3 = 5 $$  $$ \sigma_{X+Y} = \sqrt{\sigma_{X}^2 + \sigma_{Y}^2 + 2 \rho_{XY} \sigma_x \sigma_y} $$  However, I am also asked what the distribution of $ X + Y $ is, what if $X$ and $Y$ are jointly normally distributed, and what if $X$ and $Y$ are not jointly normally distributed. I am unsure how I would go about determining what the distribution of $X + Y$ is, though I assume that    If $X$ and $Y$ are jointly normally distributed then $X + Y$ is normally distributed If $X$ and $Y$ are not jointly normally distributed then $X + Y$ is not normally distributed   How do I determine what the distribution of $X + Y$ is when I am not told whether they are jointly distributed?    Related  I read the following post, but wasn't sure how to apply what was said to my current problem, if I could at all.
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I need to give the distribution of $X+Y$ and $X-Y$ knowing that $(X,Y)$ is bivariate normally distributed with marginal means 1, marginal variances 1 and correlation $p=0.2$.Is it right that the marginal mean are simply $E(X)=1$ and $E(Y)=1$ ? I don't see where I should start knowing that.
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I wish to use the  to calculate the variance of a normal-distributed function. For this, I need the expected value of $X$ as well as the one of $X^2$. Intuitively, I would have assumed that $E(X^2)$ is always equal to $E(X)^2$. In fact, I cannot imagine how they could be different.  Could you explain how this is possible, e.g. with an example?
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If I have a normal distribution with a given mean and variance and apply a logistic transform to it, what is the mean and variance of my transformed variable?   This seems like it has to be a well known problem, but I haven't been able to find a quick reference anywhere.
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Problem: This is problem 33 of section 2.6 (conditioning) in Bertsekas, Tsitsiklis intro to probability text.  We are given that a coin has probability of heads equal to p and tails equal to q and it is tossed successively and independently until a head comes twice in a row or a tail comes twice in a row. What isthe expected value of the number of tosses?  Solution:  Let $X$ be a random value that tracks the number of tosses. In addition, let $H_k, T_k$ be the event that a head or tail comes up on the kth flip respectively. The solution I am looking at involves using total expectation theorem.  I outline the following steps in the solution.  We can partition the sample space into $H_1$ and $T_1$, allowing us to invoke the total expectation theorem. Hence,  $E[X] = pE[X|H_1] + qE[X|T_1]$  Again by total expectation theorem, we can utilize another partition to obtain  $E[X|H_1] = pE[X|H_1,H_2] + qE[X|H_1,T_2] = 2p + q(1 + E[X|T_1])$  $E[X|T_1] = qE[X|T_1,T_2] + pE[X|T_1,H_2] = 2q + p(1+E[X|H_1])$  Stuck: The solution follows up by saying that we can combine the above two relations and use the fact that p+q = 1 to obtain  $E[X|T_1] = \frac{2+p^2}{1-pq} , E[X|H_1] = \frac{2+q^2}{1-pq}$  I have been staring at this for a while now, but I cannot seem to see the steps of the computation. I'd be very grateful if someone could fill in the details regarding the computation. Thank you.
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I have two independent gaussian distributions to combine and I have a doubt.  Let's say we have $X \sim N(\mu_x,\sigma^2_x)$ and $Y \sim N(\mu_y,\sigma^2_y)$. I want to mix the two variables with a 0.5 weight each, so I obtain a new variable $Z=0.5*X+0.5*Y$ with mean $\mu_z=0.5*\mu_x+0.5*\mu_y$.  Is it right to say that $\sigma^2_z=0.5^2*\sigma^2_x+0.5^2*\sigma^2_y$?  Because I clearly obtain less variance than intuitively expected when combining distributions.
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I'm trying to understand my teacher's solution to this problem:     Let $X_1,X_2,...$ sequence of iid random variables with expected value   $\mu$. Define $Y_n = \frac{2}{n(n+1)} \sum_{j=1}^{n} jX_j$. Prove that   $Y_n \to_\mathbb{P} \mu$.   His solution:  Let $\sigma^2 = \rm{Var}(X_i)$.  $E(Y_n) = \frac{2}{n(n+1)} \sum_{j=1}^{n} jE(X_n) = \mu$.  $\displaystyle\rm{Var}(Y_n) = \frac{4}{n^2(n+1)^2} \sum_{j=1}^{n} j^2 \rm{Var}(X_n) = \frac{4\sigma^2}{n^2(n+1)^2}\bigg(\frac{n(n+1)(2n+1)}{6}\bigg) = \frac{2\sigma^2}{3}\bigg(\frac{2n+1}{n(n+1)}\bigg) \leq \frac{2\sigma^2}{3}\bigg(\frac{2}{3}\bigg) = \sigma^2 ,\forall n$  Then it follows by Tchebychev's Weak Law of Large Numbers.  My doubt is in the last step. Isn't  Tchebychev's WLLN only true for uncorrelated random variables? In this case, $\mathbb{E}(Y_n Y_m) \neq \mu^2$, since they aren't independent, so $\rm{Cov}(Y_n,Y_m) \neq 0$, right?  Thanks.  Edit: As u/clarinetist noted, there's a problem in this exercise, since the WLLN states the convergence of the arithmetic mean of random variables, whose expected value isn't $\mu$.
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In the Beta distribution (used to model Bernoulli probabilities), the $\alpha$ and $\beta$ parameters can be interpreted as the number of heads$+1$ and the number of tails$+1$ seen. So, if they were both $2$, it would lean towards the coin being fair and have a maximum at $0.5$. If they are both $20$, the distribution would become even surer we're dealing with a fair coin and peak even more at $p=0.5$.  What I don't get is its behavior when $\alpha$ and $\beta$ both become $&lt;1$.   In that case, it becomes U-shaped and the density peaks at $p=0$ and $p=1$. Meaning the coin is likely to be two-sided. I know there is an intuition for this since I think I had an idea about it a long time ago. However, I've been trying to recollect all day and can't piece it together. Does anyone have an intuition?
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If $X$ is a discrete random variable with expected value $\mu$ and standard deviation $\sigma$, and $Y = X^2$, how can we describe the expected value and standard deviation of $Y$? Would expected value of $Y = \mu^2$?