Expected value of random binomial distribution variables [closed]

I have no idea how to approach this:

Let $X_i$ be a binomial distribution variable with parameters $(i,p)$ where $(0<p<1)$ for all $i=0,1,2.....$

And let $N$ be a poisson distribution with $\lambda(0<\lambda)$.

$X_0, X_1,X_2....$ are independent among them and independent on $N$.

We define $Y=\sum_{i=0}^{N}X_i$

I need to calculate the expected value of $Y$.

What I know is that $E[Y]=E[\sum_{i=0}^{N}X]=\sum_{i=0}^{N}E[X]=np$ when $X$ isbinomial distribution variable

closed as off-topic by Did, Leucippus, NCh, JonMark Perry, Claude LeiboviciJun 29 '17 at 6:54

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• Can you compute $E(Y\mid N=n)$, for every $n\geqslant0$? – Did Jun 28 '17 at 19:17

Hints:

$$E(Y) = E(E(Y|N))$$

$$E(X_i) = ip$$

$$E(N) = \lambda$$

$$E(N^2) = \lambda^2 + \lambda$$

and use linearity of expectation.

Edit: Try computing yourself first, then look at my computation.

Compute $E(Y|N)$

$$E(Y|N) = E(\sum_{i=0}^{N}X_i) = \sum_{i=0}^{N}E(X_i) = \sum_{i=0}^{N}ip = \frac{N(N+1)p}{2}$$

Now, compute $E(E(Y|N)$,

$$E\left(\frac{N(N+1)p}{2}\right) = \frac{p}{2}E(N^2+N) = \frac{p(\lambda^2+2\lambda)}{2}$$

• But In my textbook it says binomial distributionare. are you sure it can handle only with negative binomial distribution? – timi Jun 28 '17 at 19:25
• I think I misunderstood your problem. I made the correction. – Dhruv Kohli - expiTTp1z0 Jun 28 '17 at 19:28
• You think that $E(N) = \lambda$ and $E(N^2) = \lambda^2$? Hmmm... – Did Jun 28 '17 at 19:38
• @Did Corrected. Exhausted. – Dhruv Kohli - expiTTp1z0 Jun 28 '17 at 19:41
• I'm really thankfull for your help. can you show me a sketch of the start of this calculation? – timi Jun 28 '17 at 19:44