Show that we always have $Y + Z = X + 4$.

Question

Let $X$ be a Geometric random variable with parameter $p =\frac{1}{2}$. We define another random variable $Y$ in terms of $X$ as follows.

$Y = \min\{X,4\}$

Here $\min\{X,4\}$ is the minimum between the value of $X$ and $4$. For example, if $X$ takes the value $3$ then $Y = \min\{X,4\} = 3$, whereas if $X = 5$, $Y = \min\{X,4\} = 4$.

We define $Z = \min\{X,4\}$, the maximum between value $X$ and $4$. Show that we always have $Y + Z = X + 4$.

I know $E[X]$ is $2$ and $E[Y]$ is $\frac{15}{8}$

I know this statement is true and makes sense but how do I go about proving it? What approach should I take?

Answer 1

This is trivial.

If $X\le4$, $Y+Z=\min(X,4)+\max(X,4)=X+4$.

If $X\ge4$, $Y+Z=\min(X,4)+\max(X,4)=4+X$.