Suppose $X$ is a discrete random variable with probability generating function:
$G_X(\theta)$ = $2(3-\theta)^{-1}$
1) If $Y$ = $X^{2}$ write down $P(Y=k)$ for $0\leq k \leq 10$, and find $E(Y)$
Firstly i know i have to get $G_X'(\theta)$ but i don't know how to do the rest. Can someone please help me with this.
Hints:
$G_X(\theta)$ is a generating function, so expand it and look at the coefficients of powers of $\theta$ to find the probabilities of $X$ taking small values, and so of $Y$ taking the squares of these values
$G'_X( 1^- ) = E[X]$ and $G''_X( 1^- ) = E[X(X-1)]$
The pgf gives you the probabilities $p_0,p_1,\ldots$ of values $0,1,\ldots$ in the form:
$$pgf(x)=p_0+p_1\cdot x+p_2\cdot x^2+\ldots$$
It is useful, first of all, because it allows you to express an infinite list of probabilities as one (finite) closed-form expression.
Note, that $pgf(0)=p_0$. To obtain the probability $p_k$, $k\ge 1$ you need to differentiate the function $k$ times, which gives you:
$$pgf^{(k)}(x)=p_k\cdot k!+p_{k+1}\cdot \frac{(k+1)!}{1!}x+\ldots,$$
take its value at 0, and divide by $k!$: $$pgf^{(k)}(0)=p_k\cdot k!$$.
Alternatively, if you can rewrite $pgf(x)$ as series, then the coefficients will give you the probabilities immediately.
In your question you are asked about the probabilities for $Y=X^2$, for values $\le 10$. First, if $X$ takes values $0,1,2,\ldots$, then $Y$ takes possible values $0,1,4,\ldots$, and there are only four of them $\le 10$. Moreover, $\mathbb{P}(Y=k^2)=\mathbb{P}(X=k)=p_k$. Now you have everything to find those 4 values.
To find $\mathbb{E}(Y)$ you need to calculate:
$$\mathbb{E}(Y)=\sum_{k=0}^{\infty}p_k\cdot k^2,$$
do it either directly, or consider the following well-known "work-around" using pgf:
$$\mathbb{E}(Y)=\mathbb{E}(X^2)=\mathbb{E}(X(X-1))+\mathbb{E}(X)=pgf''(1)+pgf'(1).$$
