$X$ is a random variable having pmf
$c(4−x)$$I_{\{−1,2,3\}}$$(x)$, for some constant $c$. Give the value of $E(X)$.
Start at a Solution:
It is my understanding that an indicator function is $1$ when a certain criteria holds. I think the notation in the question means that the criteria holds when $x = -1$, $ x = 2$, $ x = 3$.
Then $X$ = $I_{-1}$ + $I_2$ + $I_3$ = $c(4-(-1)) + c(4-2) + c(4-3)$ = $8c$.
I don't know where I'm going with this attempted solution. I'm really confused about indicator functions.
It's not true that $X = 8c$ (that doesn't make sense - $X$ is a random variable, not a constant) but that calculation you did will help you solve the problem.
You're right that $I_{\{-1,2,3\}}(x)$ is one if $x=-1,2,$ or $3$ and zero otherwise. So what is meant is that the distribution of $X$ obeys $$ P(X = -1) = c(4--1)=5c,\\ P(X = 2) = c(4-2) = 2c,\\ P(X=3) = c(4-3) = c$$ and $P(X\ne -1,2 \;\text{or}\; 3) = 0.$
So you know that $$ 1 = P(X=-1) + P(X=2) + P(X=3) = 8c$$ and thus $c=1/8.$
Now to compute $E(X)$ you take $$ (-1)P(X=-1) + 2P(X=2)+3P(X=3).$$