I am unable to understand how the writer has derived the value of "a" in part a, and P.M.F in part "b" in the following question of "Introduction to Probability by Bertsekas":
I am completely clueless about the part "a", regarding part "b", I feel only one value of "z" satisfies the condition x^2=z, that is when x, and z both are 1, why the author has written probability z/28 + z/28?
Please elaborate both parts, thank you!
We want that $\sum\limits_{x\in \Omega}p_X(x)=1$, otherwise it contradicts the definition of a probability distribution function.
You have the only terms with nonzero $p_X(x)$ are those $x\in \{-3,-2,-1,1,2,3\}$ (as per the definition of the specific $p$)
The total will be:
$\sum\limits_{x\in\Omega} p_X(x) = \frac{(-3)^2}{a}+\frac{(-2)^2}{a}+\frac{(-1)^2}{a}+\frac{(1)^2}{a}+\frac{(2)^2}{a}+\frac{(3)^2}{a} = \frac{9+4+1+1+4+9}{a}=\frac{28}{a}$
Remembering that we wanted $a$ to be picked so that $\sum\limits_{x\in\Omega} p_X(x)=1$, that implies that $a$ must be $28$.
For calculating $E[X]$, it is $\sum\limits_{x\in \Omega} xp_X(x)$, which in this case is equal to:
$E[X] = (-3)\cdot \frac{(-3)^2}{28} + (-2)\cdot \frac{(-2)^2}{28}+(-1)\cdot \frac{(-1)^2}{28}+0+1\cdot\frac{1^2}{28}+2\cdot\frac{2^2}{28}+3\cdot\frac{3^2}{28} = 0$