I can give you an answer for the pmf of X-Y. From there |X - Y| is straightforward.
So we start with
$X \sim Bin(n_1, p_1)$
$Y \sim Bin(n_2, p_2)$
We are looking for the probability mass function of $Z=X-Y$
First note that the min and max of the support of Z must be $(-n_2, n_1)$ since that covers the most extreme cases ($X=0$ and $Y=n_2$) and ($X=n_1$ and $Y=0$).
Then we need a modification of the binomial pmf so that it can cope with values outside of its support.
$m(k, n, p) = \binom {n} {k} p^k (1-p)^{n-k}$ when $k \leq n$ and 0 otherwise.
Then we need to define two cases
- $Z \geq 0$
- $Z \lt 0$
In the first case
$p(z) = \sum_{i=0}^{n_1} m(i+z, n_1, p_1) m(i, n_2, p_2)$
since this covers all the ways in which X-Y could equal z. For example when z=1 this is reached when X=1 and Y=0 and X=2 and Y=1 and X=4 and Y=3 and so on. It also deals with cases that could not happen because of the values of $n_1$ and $n_2$. For example if $n_1 = 4$ then we cannot get Z=1 as a combination of X=5 and Y=4. In this case thanks to our modified binomial pmf the probablity is zero.
For the second case we just reverse the roles. For example if z=-1 then this is reached when X=0 and Y=1, X=1 and Y=2 etc.
$p(z) = \sum_{i=0}^{n_2} m(i, n_1, p_1) m(i+z, n_2, p_2)$
Put them together and that's your pmf.
$ f(z)=
\begin{cases}
\sum_{i=0}^{n_1} m(i+z, n_1, p_1) m(i, n_2, p_2),& \text{if } z\geq 0\\
\sum_{i=0}^{n_2} m(i, n_1, p_1) m(i+z, n_2, p_2), & \text{otherwise}
\end{cases}
$
Here's the function in R and a simulation to check it's right (and it does work.)
https://gist.github.com/coppeliaMLA/9681819
1) Knowing that in the first four hours $8$ persons got into the bank, calculate the probability of exactly half of them having entered past the first two hours.
I define the random variable $Y_i$ =number of clients in the i-th hour of the opening hours $Y_i$=number of clients in the i-th hour of the opening hours
for $i=1,2,3,4$ , since $Y_i \sim P(λ)$ and these random variables are independent, then $Y=Y_3 +Y_4 \sim P(2λ)$ . The random variable that seems suitable to calculate the probability is $X∼Bin(8,P(Y=4))$.
Fine until the last sentence. Close, but no.
What you have not taken into account is the given condition that $Y_1+Y_2+Y_3+Y_4=8$. You know that eight Poisson point events are distributed across these time intervals. But how are they distributed across these intervals?
Uniformly, that's how. Each of these four events may have occurred in the latter two hours with probability $1/2$, so the count of the events that did so is Binomially distributed, via: $X\sim\mathcal {Bin}(4, 1/2)$
$$\mathsf P(X=4) ~=~ \dbinom{8}{4}\dfrac 1 {2^8} ~=~ \dfrac {35}{128}$$
To verify the argument we work from the premises that (1) the number of clients who arrive in an interval $t$ is Poisson distributed with rate $t\lambda$, and that (2) the count of arrivals in disjoint intervals will be independent.
That is to say: If we let $X_1$ be the count of clients who arrive in the earlier two hours, and $X_2$ be the count of those who do so in the latter two hours. Then $X_1\sim\mathcal{Pois}(2\lambda)$ and independently $X_2\sim\mathcal {Pois}(2\lambda)$. Also $\mathcal X_1+X_2\sim\mathcal{Pois}(4\lambda)$
The conditional probability that $k$ clients arrived in the later two hours when given that $n$ arrived somewhen in the four hours is:
$$\begin{align}\mathsf P(X_2=k\mid X_1+X_2=n) ~=~ &\dfrac{\mathsf P(X_2=k)~\mathsf P(X_1=n-k)}{\mathsf P(X_1+X_2=k)}
\\[2ex] ~=~ & \dfrac{\dfrac{(2\lambda)^k~\mathsf e^{-2\lambda}}{k!}~\dfrac{(2\lambda)^{n-k}\mathsf e^{-2\lambda}}{(n-k)!}}{\dfrac{(4\lambda)^n\mathsf e^{-4\lambda}}{n!}}
\\[2ex] ~=~ & \dfrac{n!}{k!~(n-k)!}\dfrac 1{2^n}
\end{align}$$
Best Answer
Rather than $X\sim\operatorname{Bin}(N,p)$ I would write $X\mid N\sim\operatorname{Bin}(N,p),$ and similarly $Y\mid N\sim\operatorname{Bin}(N,1-p).$
You have \begin{align} & \Pr(X=x\ \&\ Y=y) = \operatorname E(\Pr(X=x\ \&\ Y=y\mid N)) \\[8pt] \text{and } & \Pr(X=x\ \&\ Y=y\mid N=n) = 0 \text{ unless } n = x+y. \\[8pt] \text{So } & \operatorname E(\Pr(X=x\ \&\ Y=y\mid N)) \\[8pt] = {} & \sum_{n=0}^\infty \Pr(X=x\ \&\ Y=y\mid N=n)\Pr(N=n) \\[8pt] = {} & \Pr(X=x\ \&\ Y=y\mid N=x+y)\Pr(N=x+y) \\ & \text{(All other terms in the sum vanish.)} \\[8pt] = & \Pr(X=x\mid N) \Pr(N=x+y) \\ & \text{(since conditional on $N=x+y$, the} \\ & \phantom{(}\text{events $X=x$ and $Y=y$ are the same)} \\[8pt] = {} & \binom{x+y} x p^x(1-p)^y \cdot \frac{\lambda^{x+y} e^{-\lambda}}{(x+y)!} \\[8pt] = {} & \frac{(p\lambda)^2 e^{-p\lambda}}{x!} \cdot \frac{((1-p)\lambda)^y e^{(1-p) \lambda}}{y!} \end{align} So $X,Y$ are independent Poisson-distributed random variables with expected values $p\lambda$ and $(1-p)\lambda.$