Suppose a sum of $1000$ pounds is invested with annual interest $5\%$ for $X$ years, where $X \sim \operatorname{Geom}(1/4)$. Compute the expected value of the investment’s worth at the end of the $X$ years.
I found the expected value of $X$ to be $4$, but I'm not sure what to do next?
The value after $x$ years is $1000\times(1.05^x)$. The Expected value, by definition, is $\sum_{x=1}^\infty 1000\times(1.05^x)\times P(X=x)$ For the geometric distribution with probability $p$ we have $P(X=x)=(1-p)^{x-1}p$. So the expected value is $\sum_{x=1}^\infty 1000\times(1.05^x)\times (1-p)^{x-1}p$ which can be written $1000\times1.05p\sum_{x=1}^\infty (1.05(1-p))^{x-1} $ Using the formula for the sum of an infinite geometric series (for this to be valid we need $|1.05(1-p)|<1$ we get: $1000\times1.05p\times \frac{1}{1-1.05(1-p)}$ So plug in $p=1/4$ and you should get the answer.