A random variable, X, is defined as follows. First flip a coin with a probability of p of heads. If the coin lands heads, then define X as X=0. If the coin lands tails, then X has a Poisson distribution, Poi(λ).
- Find PMF of X
For this problem I had that if the first flip was a heads, then P(X=0)=p, since I interpreted X as the number of tails. But I'm not sure about that. If the first flip was a tails, then I said it would be the poisson PMF formula: $e^{-λ}$ $λ^k$/k!
but I wasn't sure if there should be two separate PMFs or there should only be one.
- let Y~Poi(λ) and let U~Ber(p) that is independent of Y. Let Z = (1-U)Y. Show that Z and X have the same PMF.
I'm a bit lost at this problem. I know that U~Ber(p) = p and 1-p, and that Y~Poi(λ) = $e^{-λ}$ $λ^k$/k!. I'm not sure where to proceed.
- We know that E(Y) = Var(Y) = λ. Using this and part 2, find E(X) and Var(X). You can also use the fact that if V and W are independent random variables, then E(VW) = E(V)*E(W)
Again, not sure about part 2, so I'm also not sure about this problem.
There should only be one, and you should combine the disjoint cases. Your instinct here is correct.
$$\mathsf P(X=k)=\begin{cases}\mathsf P(\text{heads or, tails and Poisson count of }0)&:& k=0\\\mathsf P(\text{tails and Poisson count of }k)&:& k\in\Bbb N~, k>0\\0&:&\text{otherwise}\end{cases}$$
Part 1 and Part 2 are really just asking the same problem, with part 2 being a little more formal about it.
Not quite. More specifically, what the distributions tell you are the probability mass functions for the variables, which may be combined to derive that for $Z$
Let $U\sim\mathcal{Bern}(p)$ be the random variable indicating the count for heads.$$\mathsf P(U=\theta)=p^\theta~(1-p)^{1-\theta}\quad\big[\theta\in\{0,1\}\big]$$
Let $Y\sim\mathcal{Pois}(\lambda)$ be the poison random variable.
$$\mathsf P(Y=k)=\dfrac{\lambda^k\mathrm e^{-k}}{k!}\quad\big[k\in\Bbb N\big]$$
Then we have that $Z= (1-U)~Y$ , and so $Z=0$ when the coin shows heads, or else when it shows tails and the Poisson count is zero. Likewise, $Z$ is nonzero only when the coin shows tails and the Poisson count is nonzero too. Additionally, $U$ and $Y$ are independent.
This is exactly as is the case for $X$, so $X$ and $Z$ are identically distributed.
$$\begin{align}\mathsf P(X\,{=}\,k)&=\mathsf P(Z\,{=}\,k)\\[2ex]&=\begin{cases}\mathsf P(U\,{=}\,1\cup(U\,{=}\,0\cap Y\,{=}\,0))&:& k=0\\[1ex]\mathsf P(U\,{=}\,0\cap Y\,{=}\,k)&:&k\in\Bbb N{\smallsetminus}\{0\}\\[1ex]0 &:&\text{otherwise}\end{cases}\\[2ex]&=\begin{cases}\mathsf P(U\,{=}\,1)+\mathsf P(U\,{=}\,0)~\mathsf P(Y\,{=}\,0)&:& k=0\\[1ex] \mathsf P(U\,{=}\,0)~\mathsf P(Y\,{=}\,k)&:&k\in\Bbb N\smallsetminus \{0\}\\[1ex]0 &:&\text{otherwise}\end{cases}\\[2ex]&~~\vdots\end{align}$$
Just fill in the ellipsis.
Here you just need to know the mean and variance for Bernoulli and Poisson variables, which you were told.
$${\mathsf E(U)=p\\\mathsf {Var}(U) = p(1-p)\\\mathsf E(Y)=\lambda\\\mathsf {Var}(Y)=\lambda}$$
Everything else is just applying the definitions for expectation and variance and hint you were given. Note: since $U$ and $Y$ are independent, so too are $(1-U)$ and $Y$, and also $(1-U)^2$ and $Y^2$ so the hint tells you:
$${\mathsf{E}((1-U)Y)=\mathsf E(1-U)~\mathsf E(Y)\\\mathsf E((1-U)^2 Y^2)=\mathsf E((1-U)^2)~\mathsf E(Y^2)}$$