1. ## Discrete rv question

I am stuck on the below problem. I am not sure how to approach. What I have done so far is used the fact that $V (Y ) = V (X )$ and expanded the the variance formula to include $E(Y^{2})-E(Y)^{2}=E(X^{2})-E(X)^{2}$, but I have not had much luck with it. I also thought about settin gup a system, but I am not sure how I would do that.

Q: Suppose Y is a discrete random variable having $P(Y= -1)=1/4$, $P (Y = 0) = 1/4$,
$P (Y = 1) = 1/4$, and $P (Y = 2) = 1/4$. Consider the discrete random variable X having
$P (X =
-1) = p_{1}$
, $P (X = 0) = p_{2}$ , $P (X = 1) = p_{3}$ , and $P (X = 2) = p_{4}$ . Also, suppose
p2 = 2p1 . Find p1 , p2 , p3 , and p4 such that $E(Y ) = E(X )$ and $V (Y ) = V (X )$.

Any help would be much appreciated, thanks

2. Originally Posted by Danneedshelp
I am stuck on the below problem. I am not sure how to approach. What I have done so far is used the fact that $V (Y ) = V (X )$ and expanded the the variance formula to include $E(Y^{2})-E(Y)^{2}=E(X^{2})-E(X)^{2}$, but I have not had much luck with it. I also thought about settin gup a system, but I am not sure how I would do that.

Q: Suppose Y is a discrete random variable having $P(Y= -1)=1/4$, $P (Y = 0) = 1/4$,
$P (Y = 1) = 1/4$, and $P (Y = 2) = 1/4$. Consider the discrete random variable X having
$P (X =
-1) = p_{1}$
, $P (X = 0) = p_{2}$ , $P (X = 1) = p_{3}$ , and $P (X = 2) = p_{4}$ . Also, suppose
p2 = 2p1 . Find p1 , p2 , p3 , and p4 such that $E(Y ) = E(X )$ and $V (Y ) = V (X )$.

Any help would be much appreciated, thanks
$p_1 + p_2 + p_3 + p_4 = 1 \Rightarrow 3p_1 + p_3 + p_4 = 1$ .... (1)

You need two more equations:

Use the definition of expected value to calculate E(X) and E(Y) and hence get a second equation using E(X) = E(Y).

Use the definition of variance to calculate Var(X) and Var(Y) and hence get a third equation using Var(X) = Var(Y).

If you need more help, please show all your working and state where you're stuck.