Computing the mean of $5X$ given that $X$ is discrete and uniformly distributed

Question

I'm trying to understand the result 2 on this page by way of example. It says for a discrete random variable $X$, with PMF $f(x)$, that the PMF of the transformed random variable $Y=r(X)$ we get a PMF

$$g(y)=\sum_{x\in r^{-1}\{y\}}f(x)$$

Here's the problem for which I'm trying to apply this result.

Suppose that $X$ is uniformly distributed on the integers $0$ to $9$. What is the mean of $Y=5X$?

For starters, I know that

$$f(x)=\begin{cases}\frac1{10}&\text{for }x\in\{0,\ldots,9\}\\0&\text{otherwise}\end{cases}$$

and since $Y=r(X)=5X$, we have $X=r^{-1}(Y)=\frac Y5$. But then the theorem suggests that the new variable's PMF would be

$$g(y)=\sum_{x\in r^{-1}\{y\}}f(x)=\sum_{x=0}^9f(x)$$

The sum evaluates to $1$, so I'm convinced I'm misunderstanding something important.

What does it mean to sum over all $x\in r^{-1}\{y\}$? Is it not true that $r^{-1}\{y\}=\{0,\ldots,9\}$?

I've thought about this in another way using the CDF for $X$. Letting $F(X)$ denote the CDF for $X$ and $G(Y)$ the CDF for $Y$,

$$G(y)=\mathbb P(Y\le y)=\mathbb P(5X\le y)=\mathbb P\left(X\le\dfrac y5\right)=F\left(\dfrac y5\right)$$

I know that

$$F(x)=\begin{cases}0&\text{for }x<0\\\frac n{10}&\text{for }n-1\le x<n,~n\in\{0,\ldots,9\}\\1&\text{for }9\le x\end{cases}$$

which would yield

$$F\left(\dfrac y5\right)=\begin{cases}0&\text{for }y<0\\\frac n{10}&\text{for }5n-5\le y<5n,~n\in\{0,\ldots,9\}\\1&\text{for }45\le y\end{cases}$$

and so, if I'm not mistaken, the PMF for $Y$ is

$$g(y)=\begin{cases}\frac1{10}&\text{for }y\in\{0,5,\ldots,45\}\\0&\text{otherwise}\end{cases}$$

If this is correct, then

how do I arrive at this result using the theorem from the provided link?

Answer 1

The support of $X$, which I denote $\mathcal{X}$, is $\{0, 1, \dots, 9\}$.

The image of $\mathcal{X}$ induced by $r$ is $r(\mathcal{X}) = \{0, 5, \dots, 45\}$. These are the elements $y$ that are being referred to. Let's let $Y = \{0, 5, \dots, 45\}$.

The inverse image of $Y$ is given by $r^{-1}(Y) = \{0, 1, \dots, 9\}$, which is simply $\mathcal{X}$.

Here's how the definition $$g(y) = \sum_{x \in r^{-1}(\{y\})}f(x)$$ works. Notice that $\{y\}$ is a singleton set - i.e., $y$ is only one value.

Fix $y$. We know that $y$ can be any one of the values $\{0, 5, \dots, 45\}$.

Thus, for example, $$g(0) = \sum_{x \in r^{-1}(\{0\})}f(x)\text{.}$$

What is the inverse image of $\{0\}$ induced by $r$? It is just $\{0\}$. Hence, $$g(0) = \sum_{x \in r^{-1}(\{0\})}f(x)=f(0)\text{.}$$ Next, $$g(5) = \sum_{x \in r^{-1}(\{5\})}f(x) = f(1)$$ since $r^{-1}(\{5\}) = 1$.

Keep doing this for every one of the ten values in $Y$.

Answer 2

For each $y$ in the codomain, $r^{-1}(\{y\}),$ which can be informally written as $r^{-1}\{y\}$, is the set of things that map to $Y$. In this case, your function is

$x \mapsto 5x$

So for $y = 5$, $r^{-1}\{5\} = \{1\}$; for $y = 10$, $r^{-1}\{10\} = \{2\}$. For $y = 6$, $r^{-1}\{6\} = \{\},$ the empty set.

Now that you see that for the $y$ values you care about, the inverse image always contains exactly one point, you know htat the prob mass associated with each of $ 0, 5, 10, 15, 20, ...$ is $1/10$.

Answer 3

For your first issue, let's take a couple of values of $y$, namely $15$ and $22$, and see what happens when you turn the crank on $$ g(y)=\sum{x\in r^{-1}(y} f(x) $$ For $y=15$, $r^{-1}(15) = \{3\}$ because only $x=3$ gives $y=15$. So $g(15) = f(3) = \frac1{10}$.

For $y=22$, $r^{-1}(2) = \{3\}$ because only $x=4.4$ gives $y=22$. So $g(22) = f(4.4) = 0$.

Thus $g(y) = \frac1{10}$ for $y\in\{0,5,10,15,\ldots,45\}$ and zero otherwise.

Note that for a specific value of $y$, $r^{-1}(y)$ does not include every possible value of $x$.