Find expectation of of $max(X,5)$ from given probability mass function

Question

Let $X$ be random variable with pmf

$P(X=n)=\dfrac{1}{10},n=1,2...10$

$0$ otherwise

Find $E(max(X,5))$

Let $Y=max(X,5)$

$P(Y=y) = \begin{cases} 5, & \text{if $X \le5$ } \\[2ex] X, & \text{if $X >5$ } \end{cases}$

$E(y)=\sum yP(Y=y)$

$E(y)=E(5)+E(X)$

$=5+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}+\dfrac{10}{10}$

I am not sure if I am going in right direction.

Answer 1

In cases like this it is not necessary (and not handsome) to go for the distribution of derived random variable $\max(X,5)$.

You can just work out:$$\mathbb E\max(X,5)=\sum_{n=1}^{10}\max(n,5)P(X=n)=\frac1{10}\sum_{n=1}^{10}\max(n,5)$$