Q.4.15 - A First Course in Probability

Question

The number of eggs laid on a tree leaf by an insect of a certain type is a Poisson random variable with parameter $\lambda$ . However, such a random variable can be observed only if it is positive, since if it is $0$ , then we cannot know that such an insect was on the leaf. If we let $Y$ denote the observed number of eggs, then

$P\{Y=i\}=P\{X=i\mid X>0\}$

where $X$ is Poisson with parameter $\lambda$. Find $E\left[Y\right]$.

Step-by-Step Solution

Answer

The value of $EY=\frac{\lambda }{1-e^{-\lambda }}$

Step 1: Given information

Given in the question that , The number of eggs laid on a tree leaf by an insect of a certain type is a Poisson random variable with parameter $\lambda$. However, such a random variable can be observed only if it is positive, since if it is 0 , then we cannot know that such an insect was on the leaf. If we let $Y$denote the observed number of eggs, then

$P\{Y=i\}=P\{X=i\mid X>0\}$

where $X$ is Poisson with parameter $\lambda$.

Step 2: Explanation

We have,

$P(Y=i)=P(X=i\mid X>0)=\frac{P(X=i)}{P(X>0)}=\frac{P(X=i)}{1-P(X=0)}$

$=\frac{1}{1-e^{-\lambda }}P(X=i)$

Using the definition of expectation, we have that

$EY=\sum _{i=1}^{\infty }iP(Y=i)=\frac{1}{1-e^{-\lambda }}\sum _{i=1}^{\infty }iP(X=i)$

$=\frac{1}{1-e^{-\lambda }}\sum _{i=0}^{\infty }iP(X=i)=\frac{1}{1-e^{-\lambda }}EX$

$=\frac{\lambda }{1-e^{-\lambda }}$

Step 3:Final answer

$EY=\frac{\lambda }{1-e^{-\lambda }}$