During an experiment trial of $n$ tossed coins I have $X_1 ∼ B(n, 0.5)$. It means I have a binomial r.v. that expresses the chances of success - landing heads $(x = 1)$ as $0.5$.
When tossing again the remaining coins $R_1 = n - X_1$ we can derive $X_2$.
My question are
- Does this mean that on the second trial r.v. becomes $X_2 ∼ B(n - X_1, 0.5)$
- Also I know that capital letters represent a variable. In this case should $R_1$ be considered one? How can I find the pmf of $X_2$ while $n$ is unknown?
I'm trying to wrap my head on these concepts. Kindly point me in the right direction.
Hint:$$P(X_2=k)=\sum_{i=0}^nP(X_2=k\mid X_1=i)P(X_1=i)$$
Here $P(X_2=k\mid X_1=i)=P(Y_i=k)$ where $Y_i\sim\mathsf{Bin}(n-i,0.5)$
Note that $P(X_2=k\mid X_1=i)=0$ if $k+i>n$, so actually you only have summate for $i\leq n-k$ .
Can you work this out yourself?
edit:
$\begin{aligned}P(X_{2}=k) & =\sum_{i=0}^{n}P(X_{2}=k\mid X_{1}=i)P(X_{1}=i)\\ & =\sum_{i=0}^{n-k}P(X_{2}=k\mid X_{1}=i)P(X_{1}=i)\\ & =\sum_{i=0}^{n-k}\binom{n-i}{k}0.5^{n-i}\binom{n}{i}0.5^{n}\\ & =\sum_{i=0}^{n-k}\binom{n-i}{k}\binom{n}{i}0.5^{2n-i}\\ & =\sum_{i=0}^{n-k}\binom{n}{k}\binom{n-k}{i}0.5^{2n-i}\\ & =\binom{n}{k}0.5^{n+k}\sum_{i=0}^{n-k}\binom{n-k}{i}0.5^{n-k-i}\\ & =\binom{n}{k}0.5^{n+k}\left(1+0.5\right)^{n-k}\\ & =\binom{n}{k}0.5^{n+k}1.5^{n-k} \end{aligned} $