Find the pmf of $Y=X^2$

Question

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 $P(Y\ge3)$.

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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.

Answer 1

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.

Answer 2

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

Answer 3

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 $$P(Y\geq3)=\frac{4}{10}$$