<s>[INST] Probability distribution: finding 'a' Consider the random variable X with probability distribution,
$$
P(x) =
\begin{cases}
x^2/a & -2,-1,0,1,2 \\
0 & \text{otherwise}
\end{cases}
$$

a) Find $a$ and $E(X)$.
b) What is the probability distribution of random variable $Z = (X - E[X])^2$?
c) Using part b), compute the variance of $X$.

I think I understand parts b, c, and the second part of a, but I'm not sure how to find $a$. Maybe there's a formula I'm missing?

a) a = 10, E(X) = 0
b) $$P(Z = z) = 
 \begin{cases} 
      0 & z = 0 \\
      \frac{2}{10} & z = 1 \\
      \frac{8}{10} & z = 2 
  \end{cases}$$
alt. b) $$P(Z = z) = 
 \begin{cases} 
      \frac{2}{5} & z = 1 \\
      \frac{4}{5} & z = 4 \\
      0 & otherwise
  \end{cases}$$
 [/INST] 
a) a = 10, E(X) = 0

Yes. $
{~\\{P_X(x) =
\begin{cases} 0/a^2 & : & x=0 \\ 1/a^2 &:& x\in\{-1,1\} \\ 4/a^2 & : & x\in\{-2,2\}\\ 0 & : & \text{otherwise}
\end{cases} }\\~\\ {\therefore ~a=10, \mathsf E(X)=0}}
$


b) ${P(Z = z) = \begin{cases} 
      \frac{1}{5} & z = 0 \\
      \frac{2}{5} & z = 1,4 
  \end{cases}}$

No, but close. Try again.
Having found $a=10$ and $\mathsf E(X)=0$ you know know that $Z=X^2$ and can find $P_Z(z)= P_X(+\surd z)+ P_X(-\surd z)$  since: $$
P_X(x) =
\begin{cases} 0 & : & x=0 \\ 1/10 &:& x\in\{-1,1\} \\ 4/10 & : & x\in\{-2,2\}\\ 0 & : & \text{otherwise}
\end{cases}
$$
Therefore ...

c) Hint $\mathsf {Var}(X)=\mathsf E((X-\mathsf E(X))^2)$ and $Z=(X-\mathsf E(X))^2$ so...
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