Let X be a random variable that takes value of integers from 1 to 10 with equal probability respectively.
I am trying to compute E[X|X>4].
The attempt I made are:
$\begin{equation} \begin{split} E[X|X>4] &= \sum\limits_{n=1}^{10}nP(X=n|X>4)\\ &= \sum\limits_{n=5}^{10}nP(X=n|X>4)\\ &=\sum\limits_{n=5}^{10} n P(X=n)\\ &=\dfrac{1}{10}\sum\limits_{n=5}^{10}n\\ &=\dfrac{45}{10} \end{split} \end{equation}$
My concern is whether the step $P(X=n|X>4)$ is equivalent to $P(X=n)$ for n larger than 4 onwards?
The other sidenote question;
For $P(X=4|X=5)$, if we expand the above expression, we have by definition: $\dfrac{P(X=4,X=5)}{P(X=5)}$ and hence, can we say that the expression is zero because for the numerator, it is impossible for X to be 4 and 5 at the same time?
Let $I$ denote the indicator function, i.e. $I(A)=1$ if $A$ is true and $I(A)=0$ if $A$ is false. Then
$$E(X|X>4)= \frac{E\bigl( XI(X>4) \bigr)}{P(X>4)} = \frac{1}{6} \sum_{x=5}^{10} x=\frac{45}{6}=\frac{15}{2}.$$