Finding Expectation and Variance of $X_1$ and $X_2$

Question

Let $X_1$ and $X_2$ be random variables such that $E(X_i)=μ_i$ and $Var(X_i)=α_i^2$.
A. Find $E(X_1+X_2)$ and $E(X_1-X_2)$ in terms of the μ's and α's.
B. Suppose that $E(X_1X_2)=α$. Find $Var(X_1+X_2)$. When does it equal $Var(X_1)+Var(X_2)$?
C. Assume that $X_1$ and $X_2$ are independent. Find $Var(3X_1-2X_2)$.

Please explain how to solve these questions because I don't know where to start. Thanks.

Answer 1

This is simple problem.

A). $\mathsf{E}(X_1+X_2)=\mathsf{E}(X_1)+\mathsf{E}(X_2)=\mu_1+\mu_2$ (since expectation is linear operator, it can be shown easily by taking convolution of distribution of $X_1+X_2$. Similarly $\mathsf{E}(X_1-X_2)=\mathsf{E}(X_1)-\mathsf{E}(X_2)=\mu_1-\mu_2$

B) $var(X_1)=\mathsf{E}((X_1+X_2)-(\mu_1+\mu_2))^2=\mathsf{E}((X_1-\mu_1)+(X_2-\mu_2))^2=\mathsf{E}((X_1-\mu_1)^2+(X_2-\mu_2)^2+2(X_1-\mu_1)(X_2-\mu_2))=\alpha_1+\alpha_2+2\mathsf{E}(X_1X_2-X_1\mu_2+X_2\mu_1+\mu_1\mu_2)=\alpha_1+\alpha_2+2(\alpha-\mu_1\mu_2) $

C) Since $X_1, X_2$ are independent, hence $Var(3X_1-2X_2)=9\alpha_1+4\alpha_2$. $\square$

Answer 2

A) Let $Y$ be a linear combination of (say) two random variables $X_1$ and $X_2$. Specifically, let $Y=cX1+dX_2$ where $c$ and $d$ are constants. Then $$E(Y)=E(cX_1+dX_2)=cE(X_1)+dE(X_2).$$ In particular, $E(X_1+X_2)=E(X_1)+E(X_2)$ and $E(X_1-X_2)=E(X_1)-E(X_2)$. So the answers are $\mu_1+\mu_2$ and $\mu_1-\mu_2$. In the rest of the answer, we indicate the pieces that go into the calculation, but do not put the pieces together.

B) The variance of any random variable $W$ is $E(W^2)-(E(W))^2$. A very important fact!

Let $W=X_1+X_2$. Then $$E(W^2)=E(X_1^2+2X_1X_2+X_2^2)=E(X_1^2)+E(X_2^2)+E(2X_1X_2).$$ Now use the fact that $E(X_i^2)=\text{Var}(X_i)+(E(X_i))^2$, and that we know $E(X_1X_2)$ to finish the calculation.

C) It is a general fact that if $X_1$ and $X_2$ are independent, then the variance of $cX_1+dX_2$ is equal to $c^2\text{Var}(X_1)+d^2\text{Var}(X_2)$. For a proof, we can use the fact that the variance of $cX_1+dX_2$ is $$E((cX_1+dX_2)^2)-(E(cX_1+dX_2))^2,$$ and then calculate as in Part B. We will be using the fact that if $X_1$ and $X_2$ are independent, then $E(X_1X_2)=E(X_1)E(X_2)$.