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Let $X$ be a random variable with pmf $$p(n)=\left(\frac{3}{4}\right)^{n-1} \frac 1 4, \qquad n=1,2,\ldots$$

$$E(X-3\mid X>3)=$$ Can anybody give me the ideas how to proceed

My approach is that I find the pmf of $X-3$ which i got $P(X-3=i)=\left(\frac3 4 \right)^{2+i}\frac{1}{4}$ ,$i=1,2,\ldots$ Then i calculated the moment generating fuction The final answer i got is $\frac{27}{4}$

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    Do you know about the memorylessness of the exponential distribution? That's what you need here. – 2017-02-15

2 Answers 2

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Hint: It looks like a familiar type of distribution; one with an interesting property.

Geometric.   Memoryless .

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By definition,

$$E(X-3\mid X>3)=\sum_{i=4}^\infty (i-3)P(X=i\mid X>3)=\sum_{i=4}^\infty(i-3) \frac{P(X=i)}{P(X>3)},$$

then you can use the given pmf and calculate the sum.

Also note the following identity holds: $$\sum_{n=1}^\infty nr^{n-1}=\frac{1}{(1-r)^2},\quad r\in (0,1).$$