Expectation of the square of the total coin tosses

Question

Consider the random variable $X$ with the values defined by the following procedure. Initially, we set $x = 10$. We toss a coin (with the probability $p = 0.01$ of heads), decrease the value of $x$ by $1$ and if we get head then increase the value of $x$ by $5$. We repeat these coin tosses unless $x = 0$. The total coin tosses are values of $X$. How can $\mathbb{E}(X^2)$ be found? Are there any standards mathematical tools for the problem of such type?

Answer 1

I would go about it like this:

$X=X_1+X_2+\ldots+X_{10}$, where $X_i$ is the number of coin tosses to successively decrease your $x$ by one unit.

Now you see that $X_i$ are i.i.d., so

$E[X]=10E[X_1]$ and $\operatorname{Var}(X)=10\operatorname{Var}(X_1)=E[X^2]-(E[X])^2$

So $E[X^2]=10\operatorname{Var}(X_1)+100(E[X_1])^2$

The only thing left is to calculate $E[X_1]$ and $E[X_1^2]$ which is done by conditioning on the first throw:

$E[X_1]=p\cdot5E[X_1]+(1-p)\cdot 1$

And similarly, but more tedious for $E[X_1^2]$:

$E[X_1^2]=p\cdot(1+5E[X_1^2]+5E[X_1]+10(E[X_1])^2)+(1-p)\cdot 1^2$