Is this expected value evaluated correctly?

Question

Let $p$ be a probability of appearing lower side of coin and $q=1-p$ be a probability of appearing upper side of coin. We throw the coin to the moment $N$ of appearing the lower side twice in the last two throws. I am obliged to find expected value of random variable $N$ (for example if $N=3$ then the lower side of coin would appear in the second and third throw).

My solution:

I used of course the definition of expected value for discrete case $$EX=\sum\limits_{i=1}^{\infty} x_i p_i. $$ For random variable $N$ we have $EN=1\cdot 0 +2\cdot p\cdot p+3\cdot q\cdot p\cdot p+4\cdot 1 \cdot q\cdot p \cdot p+\ldots=\sum\limits_{i=2}^{\infty}i q p^2 .$

If this reasoning is correct is it possible to somehow evaluate the series which I recieved? I do not know if it is final solution.

Answer 1

As commented your approach is not okay.

Hint:

Let $Y$ denote the number of trials needed to arrive at the first upper side.

Then:$$\mathbb EN=$$$$P(Y=1)\mathbb E[N|Y=1]+P(Y=2)\mathbb E[N|Y=2]+P(Y>2)\mathbb E[N|Y>2]\tag1$$ Here:

• $\mathbb E[N|Y=1]=1+\mathbb EN$
• $\mathbb E[N|Y=2]=2+\mathbb EN$
• $\mathbb E[N|Y>2]=2$

(do you see why?)

Combined with the info mentioned in the bullets equality $(1)$ enables you to find $\mathbb EN$.