<s>[INST] finding Moment generating function and CDF with pmf The random variable X has the pmf f(-1)=1/4, f(0)=1/8, f(1)=1/4, f(2)=3/8
a) How would you draw the c.d.f with points (-2,F(-2)), (-1,F(-1)), (0,F(0)), (1,F(1)), (2, F(2)), (3,F(3))
b)Write the MGF of this distribution
c) Range of and p.m.f of X^(2) 
For the cdf i tried to come up with an equation that satisfies the values but couldn't I'm guessing the cdf is the antiderivative of such equation? 
for the pmf should include f(-1)=1/4 ? i thought pmf f(x)>0
 [/INST] (1) I do not think you are expected to give single "equation."
We have $F(-2)=\Pr(X\le -2)=0$.
Similarly, $F(-1)=\Pr(X\le -1)=\frac{1}{4}$.
Also, $F(0)=\Pr(X\le 0)=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$. 
Also, $F(1)=\Pr(X\le 1)=\frac{5}{8}$.
Also, $F(2)=1$. And $F(3)=1$. 
To understand what we did, remember we are calculating the cumulative distribution function. 
(2) For the moment generating function, note that we want $E(e^{Xt})$. Note that $X=-1$ with probability $\frac{1}{4}$, $X=0$ with probability $\frac{1}{8}$, $X=1$ with probability $\frac{1}{4}$, and $X=2$ with probability $\frac{3}{8}$. So $e^{Xt}$ takes values $e^{-t}$, $1$, $e^t$, and $e^{2t}$ with the probabilities just listed. The mgf is therefore
$$\frac{1}{4}e^{-t}+\frac{1}{8}+\frac{1}{4}e^{t}+\frac{3}{8}e^{2t}.$$
(3) for the mass function of $X^2$, note that $X^2$ takes on values $0$, $1$, and $4$. The event $X^2=0$ happens precisely if $X=0$ happens. The probability of this is $\frac{1}{8}$. 
The event $X^2=1$ can happen in $2$ ways, if $X=-1$ and if $X=1$. Thus the probability that $X^2=1$ is $\frac{1}{4}+\frac{1}{4}$. 
Finally, $\Pr(X^2=4)=\frac{3}{8}$. 
Remark: For the cumulative distribution function $F(x)$ of this kind of discrete distribution, the traditional integral is not of much use. In our case, $F(x)=0$ for all $x\lt -1$. Then at $x=-1$, $F(x)$ jumps to $\frac{1}{4}$. It remains at $\frac{1}{4}$ for all $x$ with $-1\le x\lt 0$. Then at $0$ it jumps to $\frac{3}{8}$, where it remains for $0\le x\lt 1$. At $1$ it jumps to $\frac{5}{8}$, then jumps to $1$ at $x=2$, and stays there for all $x\ge 2$. 
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