Finding $E[X]$ of $X~Geom(p)$ through $E[X|toss1]$ - why $E[X+1]$ and not $E[X-1]$?

Question

$X$ - the number of tosses we need to perform to get H(eads), $p(H)=p$. Here we deal with geometric distribution, for which $p_X(k)=(1-p)^{k-1}p$

There's this way to find ${\bf E}[X]$ by conditioning on the result of the first toss. It looks like this:

${\bf E}[X]={\bf E}[X|toss_{1}=H]*P(H)+{\bf E}[X|toss_{1}=T]*P(T)=$

$={\bf E}[X|toss_{1}=H]*p+{\bf E}[X|toss_{1}=T]*(1-p)$

${\bf E}[X|toss_{1}=H]=1$, because if we get H at the first toss, we stop tossing.

${\bf E}[X|toss_{1}=T]*(1-p)={\bf E}[X+1]*(1-p)$ and here comes the question:

Why do ${\bf E}[X+1]$ and not ${\bf E}[X-1]$? Assuming that we are looking for the overall expected number of tosses and knowing that 1 toss is done, don't we look for the expected value of the tosses that are left? I've already taken into account the first 'unfavorable toss', why then add it to $X$ instead of subtract?

Answer 1

No. $\mathsf{E}(X \mid \text{toss}_1 = \mathsf{T})$ is the expected number of tosses needed to get a $\mathsf{H}$, including the first toss which is $\mathsf{T}$. Since we have known the first toss is $\mathsf{T}$, in expectation we still need $\mathsf{E}(X)$ tosses. Therefore, in total, the expected number is $$ \mathsf{E}(X) + 1 = \mathsf{E}(X + 1) $$

Answer 2

The coin is forgetful. If you know the first throw was a failure then the number of throws remaining is the same as if you had never thrown the die at all.

So when you wish to calculate the total number of throws, you sum the number of failures you have recorded and the remaining throws until seeing a heads ($X$).

$$ E[X\mid \text{toss}_1 = H] = E[X] + 1. $$ The procedure you are following is a standard trick in renewal theory.