Moment of a probability mass function

Question

Let X be a random variable with the probability mass function

$P(X=n) = \left\{ \begin{array}{ll} \frac{1}{10}, \quad n= 1,2,3,...,10 \\ 0, \quad \quad \text{otherwise}. \end{array} \right.$

Then the value of $E(\text{max}[X,5]) $ equals_________

I know for first moment $E(X) = \sum_{x=1}^{10} x*P(X=x)$

That gives us the value of first moment as $ 5.5$.

How do we find the moment when max$(X,5)$ is given

Thank You..

Answer 1

Let $ Y=\max(X,5)$. To find $\mathbb E(Y)$ you firstly should find the pdf of $Y$.

$X=1 \Rightarrow Y=5 ,\quad X=2 \Rightarrow Y=5 , \quad X=3 \Rightarrow Y=5 , \quad X=4 \Rightarrow Y=5 , \quad X=5 \Rightarrow Y=5 $

$X=6 \Rightarrow Y=6 , \quad X=7 \Rightarrow Y=7 , \quad X=8 \Rightarrow Y=8 , \quad X=9 \Rightarrow Y=9 , \quad X=10 \Rightarrow Y=10$

Each transformated outcome has a probability of $\frac1{10}$. Therefore the pdf is

$$f_Y(y)=\begin{cases} \frac12, \, y=5 \\ \frac1{10}, \, y=6,7,8,9,10 \\ 0, \, \textrm{elsewhere}\end{cases}$$

Thus the expected value of $Y$ is $\mathbb E(Y)=\sum\limits_{y=5}^{10} y\cdot f_Y(y)$

Answer 2

No, your expression for the expectation is wrong; it should be $E(X) = \sum_{n=1}^{10} n \cdot P(X=n)$.

In general, $E(f(X)) = \sum_{n=1}^{10} f(n) P(X=n)$. Apply this formula with $f(x) = \max(x,5)$.