Find the pmf of $Y=X^2$ Let $X$ be a random variable with the following pmf:
$$
\begin{array}{c|ccccc}
             x& -2 & -1 & 0 & 1 & 2 &        \\ \hline
         p(x)
          &   3/10 &   3/10 &  1/10 &   2/10 &   1/10 &   
\end{array}
$$
Find the pmf of $Y = X^2$ and find $P(Y\ge3)$.

I am struggling to get the idea behind that. Even with a solid background in multivariable calculus.
I think $y=g(X)$, where $g(x)=x^2$.
$$
\begin{array}{c|ccccc}
             x& -2 & -1 & 0 & 1 & 2 &        \\ \hline
         g(x)
          &   4 &   1 &  0 &   1 &   4 &   
\end{array}
$$
$$P_Y(y) =
\begin{cases}\displaystyle
\sum_{x\in R_x:g(x)=y} P_X(x) ,  & \text{$y \in R_y$} \\[2ex]
0, & \text{otherwise}
\end{cases}$$


*

*I know that creating this table is somewhat necessary. But what is the meaning of all that? 

*Why is making a table like that is the pmf?

*I do not understand the summation sign with the range why does it make sense.


Any hint would be greatly appreciated.
 A: We have that the pmf of $X$ is$$p_X(x)=\begin{cases}
\frac{3}{10} & x=-2 \\
\frac{3}{10} & x=-1 \\
\frac{1}{10} & x=0 \\
\frac{2}{10} & x=1 \\
\frac{1}{10} & x=4
\end{cases}$$
Transforming this to get the pmf of $Y$ we get
$$p_Y(y)=\begin{cases}
\frac{3}{10} & y=4 \\
\frac{3}{10} & y=1 \\
\frac{1}{10} & y=0 \\
\frac{2}{10} & y=1 \\
\frac{1}{10} & y=4
\end{cases}$$
Notice that the probabilities remain the same.
Finally, we have repeated $y$ values so we combine them to get
$$p_Y(y)=\begin{cases}
\frac{4}{10} & y=4 \\
\frac{5}{10} & y=1 \\
\frac{1}{10} & y=0 \\
\end{cases}$$
We can now easily compute $$P(Y\geq3)=\frac{4}{10}$$
A: We want to know the probability that observing certain value of $Y$.
We know the probability of observing certain value of $X$.
Let say we want to know the probability of observing $Y=0$, since $Y=X^2$, we know it happens when $X=0$, and hence the probability of $Y=0$ is equal to the probability of $X=0$.
Suppose we want to know the probability of observing $Y=1$, what are the possible values of $X$, $X$ can be either $1$ or $-1$. Hence that is why we need to sum them up.$P(Y=1)=P(X=-1)+P(X=1)$.
The table that you created help you to find for a certain value of $Y$, what are the possible values of $X$ that corresponds to it. 
A: Hint: The cdf of $y$ is defined as
$F_Y(y) = P(Y\leq y) = P(X^2 \leq y) = P(X \leq \sqrt{y}) = F_X(\sqrt{y})$
Since this is problem is discrete:
$F_X(\sqrt{y}) = \sum_{x_i\leq \sqrt{y}} p(x_i)$
You can then derive the pmf from the cdf
