Finding $3rd$ Moment of an Indicator Function

Question

Consider a random variable X having pmf

$p_X(x)$ = $1\over{26}$$I_{\{25,26,27,...,49,50\}}(x)$. Give the value of the $3rd$ moment of $X$.

Attempted Solution:

The $3rd$ moment of $X$ = $E(X^3)$

My understanding of indicator functions is that $X$ is true, in this case, only for values from $25$ to $50$, with each having a probability of $1\over{26}$.

So would $E(X^3)$ just be $1\over{26}$$\sum_{X=25}^{50}$$X^3$ = $59$,$062.5$

Did I do this correctly?

Answer 1

My understanding of indicator functions is that $X$ is true, in this case, only for values from $25$ to $50$, with each having a probability of $1\over{26}$.

Indeed. $\quad\mathbb I_{\{26,27,...,50\}}(x) =\begin{cases}1 & :& x\in\{26,27,...,50\}\\0 &:& \text{elsewhere}\end{cases}$

So would $E(X^3)$ just be $\dfrac 1{26}\sum\limits_{X=25}^{50} X^3 = 59,062.5$

Did I do this correctly?

Yes, although we use small case version of the random variable for values.

$$\mathsf E(X^3)=\dfrac 1{26}\sum\limits_{x=25}^{50} x^3 ~=~ 59,062.5$$