# Conditional Expectations of 2 Variables

## Homework Statement

Suppose that the number of eggs laid by a certain insect has a Poisson distribution with mean $\lambda$. The probability that any one egg hatches is $p$. Assume that the eggs hatch independently of one another. Find the expected value of $Y$, the total number of eggs that hatch.

## Homework Equations

$E(Y_1) = E(E(Y_1|Y_2))$

## The Attempt at a Solution

Call $Y_1 = Y$ and $Y_2 = N$ with $N$ being the number of eggs laid.

Then $E(Y|N=n) = \sum_{y=0}^{n} y(\frac{n!}{y!(n-y)!})p^y (1-p)^{n-y} = np$
since given that $n$ eggs are laid, the number of eggs hatching has a constant probability $p$ and thus the number of eggs that hatch would be a binomial distribution.

so $E(Y) = E(E(Y|N=n)) = E(np) = pE(n) = p\sum_{n=0}^\infty ne^{-\lambda}\frac{\lambda^n}{n!}$
since the number of eggs laid is modeled by a Poisson distribution.
and then $E(Y) = p\lambda$

The book itself doesn't have an answer listed in the back of the book and I don't have a solutions manual, so I was relying on Chegg to check my answer. However, Chegg lists the answer as $(1-p)e+p$ and models $\lambda$ as a Bernoulli random variable. I don't see how making $\lambda$ a Bernoulli random variable makes sense, as nowhere in the problem do they mention $\lambda$ varies; it's always the same mean.

Is my answer incorrect and Chegg's correct? If so, why would we follow Chegg's solution, rather than my own? Or is this just one of the many problem's Chegg has a wrong answer to?

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StoneTemplePython
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2019 Award
What course / book is this associated with?

You are correct that the answer is in fact $p \cdot \lambda$. There is a more elegant way of doing this problem via Poisson splitting, though setting up as a conditional expectation is a powerful general approach.

- - - -
If you need a book with solutions to check against (perhaps half of the problems as is common), there are a lot out there. Blitzstein has a good intro book...

What course / book is this associated with?

You are correct that the answer is in fact $p \cdot \lambda$. There is a more elegant way of doing this problem via Poisson splitting, though setting up as a conditional expectation is a powerful general approach.

- - - -
If you need a book with solutions to check against (perhaps half of the problems as is common), there are a lot out there. Blitzstein has a good intro book...
The book is Mathematical Statistics with Applications, 7th Edition - Wackerly, Mendenhall, Scheaffer. I just used the conditional expectation as that's what homework set it was assigned from. I don't think we've covered Poisson splitting yet.

StoneTemplePython
Gold Member
2019 Award
The book is Mathematical Statistics with Applications, 7th Edition - Wackerly, Mendenhall, Scheaffer. I just used the conditional expectation as that's what homework set it was assigned from. I don't think we've covered Poisson splitting yet.
Your approach works fine. If it were me, doing this via conditioning I would have calculated

$E(Y|N=n) = E\big[ \mathbb I_1 + \mathbb I_2 + ...+ \mathbb I_n\big] =E\big[ \mathbb I_1\big] + E\big[\mathbb I_2\big] + ...+ E\big[\mathbb I_n\big] = p + p + ... + p = np$

i.e. rather than using binomial distribution directly, decompose the result into a sum of coin tosses (indicator random variables) and then apply linearity of expectations, and then do the second leg as you have.
- - - -
with respect to the Chegg 'solution', an interesting check is to rerun the exact same argument on the complementary portion that don't hatch with probability $q := (1-p)$

which gives

$(1-q)e+q$

we could associate this with a random variable $Z$ if you like.

The point is that we have a random variable $X$ that is poisson distributed with mean $\lambda$ but

$X = Y + Z$

In general for a decomposition like this $Y$ and $Z$ could be dependent, but expectations are linear, so we don't need to contemplate dependencies. Using the chegg solution you get

$E\big[X\big]= E\big[Y + Z\big] = E\big[Y\big[ + E\big[Z\big] =(1-p)e+p + (1-q)e+q = e + 1$

but in general you have

$\lambda = E\big[X\big] \neq e + 1$

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Suppose that the number of eggs laid by a certain insect has a Poisson distribution with mean $\lambda$. The probability that any one egg hatches is $p$. Assume that the eggs hatch independently of one another. Find the expected value of $Y$, the total number of eggs that hatch.

## Homework Equations

$E(Y_1) = E(E(Y_1|Y_2))$

## The Attempt at a Solution

Call $Y_1 = Y$ and $Y_2 = N$ with $N$ being the number of eggs laid.

Then $E(Y|N=n) = \sum_{y=0}^{n} y(\frac{n!}{y!(n-y)!})p^y (1-p)^{n-y} = np$
since given that $n$ eggs are laid, the number of eggs hatching has a constant probability $p$ and thus the number of eggs that hatch would be a binomial distribution.

so $E(Y) = E(E(Y|N=n)) = E(np) = pE(n) = p\sum_{n=0}^\infty ne^{-\lambda}\frac{\lambda^n}{n!}$
since the number of eggs laid is modeled by a Poisson distribution.
and then $E(Y) = p\lambda$

The book itself doesn't have an answer listed in the back of the book and I don't have a solutions manual, so I was relying on Chegg to check my answer. However, Chegg lists the answer as $(1-p)e+p$ and models $\lambda$ as a Bernoulli random variable. I don't see how making $\lambda$ a Bernoulli random variable makes sense, as nowhere in the problem do they mention $\lambda$ varies; it's always the same mean.

Is my answer incorrect and Chegg's correct? If so, why would we follow Chegg's solution, rather than my own? Or is this just one of the many problem's Chegg has a wrong answer to?
I am not familiar with Chegg's book (or whatever it is), but YOU are correct. In fact, you might like to prove a more useful result: $Y$ is itself a Poisson random variable with mean $\lambda p.$