Suppose that $(X_1, X_2, \dots)$ are independent identically distributed binary variables that take on the values $0$ and $1$ with probability $P[X_i = 1] = p$, $0 < p < 1$.
We take a new r.v., $Y$ such that the event ${Y = k}$ occurs when three consecutive successes appear at the first time. That is to say, $X_{k−2} = X_{k−1} = X_k = 1$,
We are given that there is no previous occurrence of this type. Using conditional expectations, we wish to show that $E[Y] = \frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}$.
Admittedly, conditional expectation is something that really confuses me. But I know the general formula for a discrete case is:
$E[Y|X=x]=\sum_{y}yf(y|x)$.
Furthermore, I recognize the distribution as being binomial, and the binomial distribution has $E[X]=np$ for $P[X=x]=(nCx)p^x(1-p)^{n-x}$. I just need some help putting things together. Any hints, help, solutions appreciated.
I do not have enough rep to comment (I will take this down in a few minutes if it isn't helpful)... However, it seems that $E(Y)$ should not be $(1/p)+(1/p^{2})+(1/p^{3})$ given your definition of $Y$. However, I can help with understanding the conditional probability aspect. Since you are given that there are no occurences of three consecutive $X$'s being 1, we have the following three cases to consider:
(1) $X_{k-3}=0$, in which case you need $X_{k-2}=X_{k-1}=X_{k}=1$ which occurs with probability $(1/p^{3})$
(2) $X_{k-3}=0, X_{k-2}=1$, in which case you need $X_{k-1}=X_{k}=1$ which occurs with probability $(1/p^{2})$
(3) $X_{k-3}=0, X_{k-2}=1, X_{k-1}=1, $, in which case you need $X_{k}=1$ which occurs with probability $(1/p)$
The expected value of $Y$ however, is the sum over all $k$ of this type $\sum_{k\geq2}$$k\cdot[$$P(Y=k|case(1))$+$P(Y=k|case(2))$+$P(Y=k|case(3))]$
Hope this helps