- Let $X$ be a random variable that follows a Poisson distribution of parameters $m$ and let $Y$ be random variable which the conditioned law with $X = n$ follows a binomial distribution with parameters $n,p$.
Prove that: $$ p(Y = k) = \frac{(pm)^k e^{-mp}}{k!} $$
and deduce the nature of $Y$.
During a class, a bored student watches from window the leaves falling from a tree, we admit the number of the leaves fallen from the tree by the end of the class is a random variable $X$ that follows a Poisson distribution of parameter $\lambda$:
$$p(X = k) = \frac{\lambda^k}{k!}e^{-\lambda}$$
From the hypotheses above, Why would we conclude that : $e^\lambda = \sum^{+ \infty}_{k = 0} \frac{\lambda^k}{k!}$
Compute $E(X)$ and $V(X)$.
Everytime a leaves falls on the ground, the student flips a coin which gives tails with a probability $p$ and heads with probability $q = 1 – p$, $p \in (0,1)$.
We note $T$ and $H$ the number of tails and heads obtained respectively.
3.1. For $k$ fixed, explain why the distribution of $T$ knowing $X = k$ is a binomial distribution and deduce the expression of $ p(X = k|H = a)$.
3.2 For $(a,n) \in \mathbb{N}^2$, compute $p(X = k, H = a)$.
3.3 Deduce the law of $H$ and its expected value.
3.4 Without computation, determine the law $T$.
3.5 Prove that $H$ and $T$ are independents.
EDIT
After the help of David K, I decided to edit the post and write the answers of the questions I have been stuck with.
3.2. Computing $p(H = a, X = k)$ :
We know that:
$ p(H = a| X = k) = \frac{p(H = a, X = k)}{p(X = k)} $
$X$ follows a Poission distribution of parameter $\lambda$, we also know that the number of heads from $k$ flips is a binaomial random variable, so we will have:
$$
\begin{align} p(H = a, X = k) & = p(X = k). p(H = a| X = k) \\
& = \frac{\lambda^k}{k!}e^{-\lambda}C^{a}_k p^k (1-p)^{k – a}
\end{align}
$$
3.3. Applying the result of question 1, we get:
$$p(H = a) = \frac{(q\lambda)^{a}}{a!}e^{-q\lambda} $$
$H$ follows a Poisson distribution of parameter $(q\lambda)$, its expected value is $(q\lambda)$.
3.4. T is the number of tails from $k$ flips, it is also a binomial distribution, applying the result of question 1 again, with the probability of $T$ is $p$, we get :
$$ p(T = b) = \frac{(p\lambda)^{b}}{b!}e^{-p\lambda} $$
3.5. $H$ and $T$ are independents:
$$
\begin{align}
p(H = a, T = b) & = p(H = a, X = a + b) \\
& = p(H = a| X = a + b).p(X = a + b) \\
& = C^{a}_{a+b} p^a (1-p)^{a+b-a}. \frac{\lambda^{a+b}}{(a+b)!}e^{-\lambda} \\
& = \frac{(a + b)!}{a! b!} p^a q^b \frac{\lambda^a \lambda^b}{(a+b)!} e^{-\lambda( p + q )} \\
& = \frac{(q\lambda)^a}{a!} e^{- q\lambda} . \frac{(p\lambda)^b}{b!} e^{- p\lambda} \\
& = p(H = a).p(T = b)
\end{align}
$$
Best Answer
For 3.2, you are asked to find $P(X=k, H=a),$ not $P(X=a,H=a).$ I think if you account for the $k$ and $a$ more carefully you will get a different answer than currently shown at the time I write this.
For 3.3, it is true that you can apply the result of problem 1, but for $X$ the parameter of the Poisson distribution is $\lambda$ instead of $m$ (it looks like you understood--or guessed--that part correctly) and when $X=n$ then $H$ has a binomial distribution with parameters $n,p$ -- the same statement that is made about $Y$ in problem 1. So you are correct to substitute $\lambda$ for $m,$ but you should not substitute anything for $p$ (or to be pedantic, you should substitute the variable named $p$ from problem 3 for the variable named $p$ in the problem 1, but the effect on the notation is the same).
Also for 3.3, you can either write $P(H=k)$ so you are using the variable $k$ in the exact same way as in problem 1, or you can write $P(H=a)$ and substitute $a$ where $k$ was used in problem 1. Or you can use some other variable instead of either $k$ or $a$, as long as you don't write $P(H=\lambda)$ or $P(H=p)$ (since in those cases you would be using one name for two different things).
For 3.4, think! Is it possible that $H$ and $T$ are the same, and that $H=a$ and $T=a$? Is it possible that neither $H$ nor $T$ is equal to $a$, for example $H=a-1$ and $T=a+1$? Then by what logic can you possible say that $P(T=a) = 1 - P(H=a)$?
For 3.5, you have the wrong notation. $P(H.T)$ doesn't mean anything. What you want to show is $$ P(H = a, T = b) = P(H=a) P(T=b). $$
The event $H=a, T=b$ is exactly the same event as $X=a+b, H=a.$ So if you have found the correct formulas for $P(X=k,H=a),$ for $P(H=a),$ and for $P(T=b),$ this part of the question almost answers itself.