1. ## Poisson independant floods

Number of floods in a given decade is a possoin r.v. with $\lambda > 0$. Also every flood has a chance $q$ of being deadly(independent of the other floods). What is the mean and variance of the number of deadly floods in a decade?

Proposed solution:
I know the expected number of floods in a decade is $\lambda$. Each has a chance q of being deadly. So the number of deadly floods is a binomial distribution with mean $\mu = nq$ with $n = \lambda$ so we get a mean of $\lambda q$ and variance $\lambda q (1 - q)$

My problem is with the step of taking $n = \lambda$, I don't know whether that is valid because then I'm sort of taking expectations twice. Basically I'm not sure whether my logic was correct.

Thanks in advance for the help

2. Originally Posted by utopiaNow
Number of floods in a given decade is a possoin r.v. with $\lambda > 0$. Also every flood has a chance $q$ of being deadly(independent of the other floods). What is the mean and variance of the number of deadly floods in a decade?

Proposed solution:
I know the expected number of floods in a decade is $\lambda$. Each has a chance q of being deadly. So the number of deadly floods is a binomial distribution with mean $\mu = nq$ with $n = \lambda$
This isn't correct; what would be correct is: given the (random) number $N$ of floods in a decade, the number $N_d$ of deadly floods is a binomial random variable of parameters $N$ and $q$.

Then we can deduce the mean and variance of $N_d$ from the mean and variance given $N$ and the law of $N$:

$E[N_d]=E[E[N_d|N]]=E[Nq]=E[N]q=\lambda q$
and
$E[N_d^2]=E[E[N_d^2|N]]=E[Nq+N(N-1)q^2]$ $=E[N]q+E[N(N-1)]q^2=\lambda q+\lambda^2 q^2$
(thanks to computations with the binomial and Poisson distribution), hence ${\rm Var}(N_d)=\lambda q$

OR: In fact, you can prove (or perhaps you already know) that $N_d$ is a Poisson random variable of parameter $\lambda q$, so that the above is immediate. A possible proof uses generating functions... Otherwise, if you know Poisson processes, this is a standard property: if we extract points independently with probability $q$, we get another Poisson process with intensity $\lambda q$.

3. Thankyou Laurent for your detailed explanation. Really helped my understanding. I'm interested in the proof you mentioned. ie. Proving that $N_d$ is a poisson r.v. with parameter $\lambda q$. The only generating function I've learned is the moment generating function. And I don't have a good enough conceptual understanding what exactly the m.g.f is other than the definition I've been given to really use it in a proof. So is there a link where they prove this that I could look at? Or perhaps if you have time you could outline the basic principles of such a proof?

Thanks again for the explanation.

4. Originally Posted by utopiaNow
Thankyou Laurent for your detailed explanation. Really helped my understanding. I'm interested in the proof you mentioned. ie. Proving that $N_d$ is a poisson r.v. with parameter $\lambda q$. The only generating function I've learned is the moment generating function. And I don't have a good enough conceptual understanding what exactly the m.g.f is other than the definition I've been given to really use it in a proof. So is there a link where they prove this that I could look at? Or perhaps if you have time you could outline the basic principles of such a proof?

Thanks again for the explanation.
I'd rather give you the sketch of a more direct proof ('cause it's easier to explain...):

We need to find $P(N_d=n)$, and we know both $P(N=n)$ (Poisson distribution with parameter $\lambda$) and $P(N_d=n|N=m)$ (Binomial distribution with parameter $(m,q)$). To that aim, let's decompose the event $\{N_d=n\}$ depending on the value of $N$. Of course, $N$ must be larger than $n$. We have:

$P(N_d=n)=\sum_{m=n}^\infty P(N_d=n\mbox{ and }N=m)=\sum_{m=n}^\infty P(N_d=n|N=m)P(N=m)$.

Let replace by the known quantities:

$P(N_d=n)=\sum_{m=n}^\infty {m\choose n}q^n (1-q)^{m-n} e^{-\lambda}\frac{\lambda^m}{m!}$.

Now, write ${m\choose n}=\frac{m!}{n!(m-n)!}$ so that the $m!$ simplifies out, and the term in the sum depend not on $m$ but rather on $m-n$ (write $\lambda^m=\lambda^n\lambda^{m-n}$ also). Since the sum begins at $n$, change the index into $m'=m-n$ (and put the terms independent of $m$ outside). You should recognize the series expansion of the exponential function, and finally get $P(N_d=n)=e^{-\lambda q}\frac{(\lambda q)^n}{n!}$. This is what you wanted.

5. Originally Posted by Laurent
This isn't correct; what would be correct is: given the (random) number $N$ of floods in a decade, the number $N_d$ of deadly floods is a binomial random variable of parameters $N$ and $q$.

Then we can deduce the mean and variance of $N_d$ from the mean and variance given $N$ and the law of $N$:

$E[N_d]=E[E[N_d|N]]=E[Nq]=E[N]q=\lambda q$
and
$E[N_d^2]=E[E[N_d^2|N]]=E[Nq+N(N-1)q^2]$ $=E[N]q+E[N(N-1)]q^2=\lambda q+\lambda^2 q^2$
(thanks to computations with the binomial and Poisson distribution), hence ${\rm Var}(N_d)=\lambda q$

OR: In fact, you can prove (or perhaps you already know) that $N_d$ is a Poisson random variable of parameter $\lambda q$, so that the above is immediate. A possible proof uses generating functions... Otherwise, if you know Poisson processes, this is a standard property: if we extract points independently with probability $q$, we get another Poisson process with intensity $\lambda q$.
I don't understand " $E[N_d^2]=E[E[N_d^2|N]]=E[Nq+N(N-1)q^2]$ ". Would you please explain how it works? Thanks a lot!!

6. Originally Posted by cm917
I don't understand " $E[N_d^2]=E[E[N_d^2|N]]=E[Nq+N(N-1)q^2]$ ". Would you please explain how it works? Thanks a lot!!
The second equality is because if $X$ is binomial with parameter $(n,q)$ then $E[X^2]={\rm Var}(X)+E[X]^2=Nq(1-q)+(Nq)^2$. (I factorized it another way in my post because I did the computation differently).

7. Originally Posted by Laurent
This isn't correct; what would be correct is: given the (random) number $N$ of floods in a decade, the number $N_d$ of deadly floods is a binomial random variable of parameters $N$ and $q$.

Then we can deduce the mean and variance of $N_d$ from the mean and variance given $N$ and the law of $N$:

$E[N_d]=E[E[N_d|N]]=E[Nq]=E[N]q=\lambda q$
and
$E[N_d^2]=E[E[N_d^2|N]]=E[Nq+N(N-1)q^2]$ $=E[N]q+E[N(N-1)]q^2=\lambda q+\lambda^2 q^2$
(thanks to computations with the binomial and Poisson distribution), hence ${\rm Var}(N_d)=\lambda q$
Originally Posted by Laurent
if is binomial with parameter then .
I still don't get it...
In this question, we need to find Var (Flood)
let C be the event that the earthquake is catastrophic..with prob. q, . N be the event that has earthquake with parameter n
I got E[C] = E(E[C|N]) = nq
Var (C) = E [ VAR (C|N)] + VAR (E[C|N])
VAR (C|N) = nq(1-q) ...E(C|N) = nq
--> VAR(C) = E[nq(1-q)] + VAR (nq) = nq(1-q) + n^2q^2
whis is not VAR(c)=nq

Would you please tell me which step I got wrong? Thanks!!

8. Originally Posted by cm917
I still don't get it...
In this question, we need to find Var (Flood)
let C be the event that the earthquake is catastrophic..with prob. q, . N be the event that has earthquake with parameter n
I got E[C] = E(E[C|N]) = nq
Var (C) = E [ VAR (C|N)] + VAR (E[C|N])
VAR (C|N) = nq(1-q) ...E(C|N) = nq
--> VAR(C) = E[nq(1-q)] + VAR (nq) = nq(1-q) + n^2q^2
whis is not VAR(c)=nq

Would you please tell me which step I got wrong? Thanks!!
You can do it this way indeed, except that ${\rm Var}(C|N)=Nq(1-q)$ and $E[C|N]=qN$ (given $N$, $C$ is binomial with parameter $(N,q)$). By the way, what would be n in your computation? This gives

${\rm Var}(C)=E[{\rm Var}(C|N)]+{\rm Var}(E[C|N])=E[Nq(1-q)]+ {\rm Var}(Nq)$ $=\lambda q(1-q)+\lambda q^2=\lambda q$.

(I used ${\rm Var}(Nq)=q^2{\rm Var}(N)$)