Expectation value of a geometric random variable There is a geometric random variable which has the pmf
$$p(X=n) = \left(\frac{3}{4}\right) ^{n-1}\left(\frac{1}{4}\right)  n=1,2...$$
We have to find the expectation value
$$E(X-3|X>3)$$
Intuitively I understand that this amounts to finding $$\sum_{i=1}^{\infty} i*p(X=i)$$ but I want to show this mathematically using conditional expectation value
$$E(X|A)=\sum_{x} xP(X=x|A) $$
This is what I tried
\begin{align}E(X-3|X>3)&= \sum_{x} x\frac{P(X-3 \cap X>3)}{P(X>3)}\\
                      \end{align}
For values of x >3 , P(X>3) will become 1 but I dont understand how to handle the cases when x<3 as the numerator and denominator both become 0 then.
 A: You started off in the right direction, but be sure not to mix up the random variable $X$ and a particular outcome $x$.
Recapping what you did
To simplify the notation (and avoid invoking LOTUS), I'm going to use the linearity of the expectation operator:
$$
\mathbb{E}[X - 3 \mid X > 3] = \mathbb{E}[X \mid X > 3] - 3
\text{.}
$$
Now let's tackle $\mathbb{E}[X \mid X > 3]$. We can subtract 3 later.
You correctly note that
$$
\mathbb{E}[X \mid X > 3] = \sum_{x=1}^{\infty} x \cdot P(X=x \mid X > 3)
\text{.}
$$
Now, by definition of conditional probability, we rewrite this as
$$
\mathbb{E}[X \mid X > 3] = \sum_{x=1}^{\infty} x \cdot \frac{P(X=x, X > 3)}{P(X > 3)}
\text{.}
$$

So far, that's basically what you already did. Now, let's handle the confusing parts.
Computing $P(X>3)$

For values of x >3 , P(X>3) will become 1

This isn't true, though. I think you mixed up $P(X>3)$ and $P(x >3)$. The value we care about is $P(X>3)$, which is independent of the particular value that $x$ takes on.
$$
\begin{align}
P(X > 3) =& 1 - P(X \leq 3) & \textrm{by axioms of probability}  \\
=& 1 - \sum_{x'=1}^3 P(X=x') & \textrm{for discrete distribution on positive ints} \\
=& 1 - \sum_{x'=1}^3 \left(\frac{3}{4}\right)^{x'-1}\left(\frac{1}{4}\right) & \textrm{geometric pmf} \\
=& \frac{27}{64}
\text{.}
\end{align}
$$
I could also have evaluated $P(X > 3)$ directly to compute this, but the sum over 3 terms was easier. Either way, now we have our denominator!
$$
\mathbb{E}[X \mid X > 3] = \sum_x x \cdot \frac{P(X=x, X > 3)}{P(X > 3)}
= \sum_x x \cdot \frac{P(X=x, X > 3)}{\frac{27}{64}}
\text{.}
$$
Computing $P(X=x, X > 3)$
The numerator does depend on the current value of $x$. In effect, we have a piecewise operation here:
$$
P(X=x, X > 3) = \begin{cases}
    P(X=x) & x > 3 \\
    0 & x \leq 3
\end{cases}
$$
(Why? Use the chain rule. $P(X=x, X > 3) = P(X=x) \times P(X>3 \mid X=x)$.)
Combining the numerator and denominator
Sticking this into the equation from before can be ugly, notation-wise. But we see that for $x \in \{1, 2, 3\}$, the summand is equal to 0. We can now break the sum into two parts. Rewrite:
$$
\begin{align}
\mathbb{E}[X \mid X > 3] &= \sum_{x=1}^{\infty} x \cdot \frac{P(X=x, X > 3)}{P(X > 3)} \\
&= \sum_{x=1}^{3} 0 + \sum_{x=4}^{\infty} \left[ x \cdot \frac{P(X=x, X > 3)}{P(X > 3)} \right] \\
&= \sum_{x=4}^{\infty} \left[ x \cdot \frac{P(X=x, X > 3)}{P(X > 3)} \right] \\
&= \sum_{x=4}^{\infty} \left[ x \cdot \frac{\left(\frac{3}{4}\right)^{x-1}\left(\frac{1}{4}\right)}{\dfrac{27}{64}} \right]
\text{.}
\end{align}
$$
That's easy enough to evaluate. Now just remember to subtract 3, like we said we'd do at the beginning.
