Probability with Poisson distribution and mean

Question

The number of eggs that a spider puts in a tree is distributed $Poisson(λ)$, however, this variable can only be observed if it is positive because if it is $0$ you can't know if the spider was in the tree or not. Let Y be the observed number of eggs, therefore: $P(Y=i)=P(X=i|X>0)$

Determine the E[Y].

My attempt:

I define a random variable $X$ that tells us the number of eggs that the spider puts into the tree.

So $Y=X$ iff $X>0$ and $Y=0$ in other case.

Therefore, $E(Y) = \sum_{y=1}^{\infty} (yλ^{y}e^{-λ})/y!= e^{-λ}(λ+(λ^2/2) + ...) = λe^{-λ}e^{λ} = λ$

Am I right?

Answer 1

You have already defined the correct $Y$ when you wrote $Y$ in terms of $X \sim Poisson(\lambda)$:

$$P(Y=i) = P(X=i \mid X > 0)$$

Perhaps you meant for this formula to be valid for any $i>0$, and that is certainly true. However, in fact this formula is also valid for $i=0$, because in that case $P(Y=0)= P(X = 0 \mid X > 0) = 0$.

Now expand for any $i>0$:

$$P(Y=i) = P(X = i \mid X > 0) = {P(X = i \cap X> 0) \over P(X>0)} = {P(X = i) \over P(X>0)} $$

Using the fact $X \sim Poisson(\lambda)$, you can calculate the numerator $P(X=i)$ and the denominator $P(X>0) = 1 - P(X=0)$, so that gives you $P(Y=i)$ for $i>0$.

Can you calculate $E[Y]$ from here? Further hint: $P(X>0)$ will turn out to be a common term you can "factor out" in the summation.

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BTW, doing it the above way, i.e. finding each $P(Y=i)$ and then summing, is actually the hard way for this problem. I am going through all these details mainly as a way to help you "debug" your approach. For the much easier way, see the answer by @herbsteinberg

Answer 2

Your calculation does not take into account the fact that some trees were not visited. Let $s$ be the probability that a tree was visited, then $E(Y)=s\lambda$, since the rest of the contribution to $Y=0$ has probability $1-s$.

Answer 3

There's a slight trap in this question, in that the correct answer requires you introduce a quantity that's not explicitly given in the question—namely, the probability that "the" spider was in the tree. Denote this event by $\ S\ $ and let $\ p_s=P(S)\ $. Then \begin{align} P(Y=i)&=\cases{1-p_s + p_sP(X=0\,\vert S)&if $\ i=0\ $,\\ p_sP(X=i\,\vert\,S)&if $\ i>0\ $.}\\ &=\cases{1+p_s\left(e^{-\lambda}-1\right)&if $\ i=0\ $,\\ \frac{p_s\lambda^ie^{-\lambda}}{i!}&if $\ i>0\ $.} \end{align} and so \begin{align} E(Y)&=\sum_{i=1}^\infty \frac{ip_s\lambda^ie^{-\lambda}}{i!}\\ &=p_s\lambda\ . \end{align} Personally, I find the wording of the question, as given, somewhat problematic. The fact that it mentions "the" spider suggests that you only need to consider whether a single spider was present or not. If $\ p_s\ $ is quite small it's possibly reasonable to ignore the alternative that more than one spider has been laying eggs, but otherwise, it's probably unrealistic to do so.

Thus, if this question were to arise in real life from a client investigator seeking assistance from a professional mathematician or statistician, the latter would, in my opinion, have an obligation to clarify the actual situation being investigated before attempting to provide any answer.