Expected value of sum of two dependent Binomial variables

Question

Question: Assume we toss a coin in two trials. In the first trial we toss it n times. In the second trial we toss it as many as the number of tails observed in the first trial.

Calculate the expectation of total number of tails in both trails:

Solution: This is what I have tried but can't find any closed form for the answer when n is not given. I am not sure if I have taken the right approach:

$T_1$ = Number of tails in the 1st trial
$T_2$ = Number of tails in the 2nd trial

T1 ~ Binomial(n,p)

T2 ~ Binomial(T1,p)

$E(T1 + T2) = E(T1) + E(T2) = \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n p+\sum_{i=0}^{n} E\left(T_{2} \mid T_{1}=i\right) P\left(T_{1}=i\right) =\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n p+\sum_{i=0}^{n} i p \times\left(\begin{array}{l}n \\ i\end{array}\right) p^{i}(1-p)^{n-i}$

I couldn't find any closed form for the answer. Any advice on how to approach this?

Answer 1

To calculate $E(T_2)$, consider using the law of total expectation.

$$E(X) = E\left[E(X|Y) \right]$$

Which in your case would give

\begin{align*} E[T_2] &= E\left[E(T_2|T_1) \right] \\ &=E\left[p \cdot T_1 \right] = p E\left[T_1 \right] \\ &= np^2 \end{align*}

Answer 2

Let's consider an equivalent experiment. We have $n$ identical coins, numbered from $1$ to $n$. We toss each of the $n$ coins once; then each coin which came up tails is tossed one more time. Let $X_i$ be the number of times the $i^\text{th}$ coin comes up tails, and $X=\sum_{i=1}^nX_i$ the total number of tails obtained. $$E(X)=\sum_{i=1}^nE(X_i)=\sum_{i=1}^n(p+p^2)=n(p+p^2)$$