Question

Suppose that $X_i,i=1,2,3$, are independent Poisson random variables with respective means ${\lambda }_i,i=1,2,3$. Let $X=X_1+X_2$and $Y=X_2+X_3$. The random vector $X,Y$ is said to have a bivariate Poisson distribution.
($a$) Find $E\left[X\right]$and $E\left[Y\right]$.
($b$) Find $Cov(X,Y)$.
($c$) Find the joint probability mass function $P\{X=i$, $Y=j\}$.

Short answer

($a$)Therefore, the required $E\left[X\right]$and $E\left[Y\right]$is$E\left[X\right]={\lambda }_1+{\lambda }_2;E\left[Y\right]={\lambda }_2+{\lambda }_3$

($b$)Therefore, the required $Cov(X,Y)={\lambda }_2$

($c$)Therefore,

$P\{X=i,Y=j\}=\sum _{k=0}^{\min (i,j)}{e}^{-{\lambda }_1}\frac{{\lambda }_1^{i-k}}{(i-k)!}{e}^{-{\lambda }_1}\frac{{\lambda }_3^{j-k}}{(j-k)!}\frac{{\lambda }_2^k}{k!}{e}^{-{\lambda }_2}$

Step-by-step solution

Concept Introduction(part a)

The question asks to find $E\left[X\right]$. It is given that $X_i,i=1,2,3$, are independent Poisson random variables with respective means ${\lambda }_i$, for all $i=1,2,3$.

By using the common property of expectation values, one has that:
$E\left[X\right]=E\left[X_1+X_2\right]$

$=E\left[X_1\right]+E\left[X_2\right]$

$={\lambda }_1+{\lambda }_2$

Step2:Explanation(part a)

Similarly, for $Y$ :
$E\left[Y\right]=E\left[X_2+X_3\right]$

$=E\left[X_2\right]+E\left[X_3\right]$

Therefore, the required $E\left[X\right]$and $E\left[Y\right]$is $E\left[X\right]={\lambda }_1+{\lambda }_2;E\left[Y\right]={\lambda }_2+{\lambda }_3$

Step3:Final Answer

Therefore, the required $E\left[X\right]$and $E\left[Y\right]$is $E\left[X\right]={\lambda }_1+{\lambda }_2;E\left[Y\right]={\lambda }_2+{\lambda }_3$

:Concept Introduction(part b)

Compute, $Cov(X,Y)$

:Explanation(part b)

Compute, $Cov(X,Y)$
$Cov(X,Y)=Cov\left(X_1+X_2,X_2+X_3\right)$

$=Cov\left(X_2,X_2\right)$

Explanation(part b)

$Cov(X,Y)={\lambda }_2$From the property of variance and covariance, $Cov\left(X_2,X_2\right)=Var\left(X_2\right)$

Thus,
$Cov(X,Y)=Cov\left(X_2,X_2\right)$

$Cov(X,Y)=Var\left(X_2\right)$

$Cov(X,Y)={\lambda }_2$

Final Answer(part b)

Therefore, the required $Cov(X,Y)={\lambda }_2$

Concept Introduction(part c)

Find the joint probability mass function $P\{X=i,Y=j\}$.

Explanation(part c)

Find the joint probability mass function $P\{X=i,Y=j\}$.
In order to find the joint probability function, one conditions on $X_2$, and use the property of Poisson random variables.
Accordingly:

$P\{X=i,Y=j\}=\sum _{k}P\left\{X=i,Y=j\mid X_2=k\right\}P\left\{X_2=k\right\}$

$=\sum _{k}P\left\{X_1=i-k,X_3=j-k\mid X_2=k\right\}{e}^{-{\lambda }_2}\frac{{\lambda }_2^k}{k!}$

:Explanation(part c)

Since
$X=X_1+X_2$and $Y=X_2+X_{3}Y=X_2+X_3$

$=\sum _{k}P\left\{X_1=i-k,X_3=j-k\right\}{e}^{-{\lambda }_2}\frac{{\lambda }_2^k}{k!}$

$=\sum _{k}P\left\{X_1=i-k\right\}P\left\{X_3=j-k\right\}{e}^{-{\lambda }_2}\frac{{\lambda }_2^k}{k!}$

:Explanation(part c)

It is given that $X_i,i=1,2,3$ are independent Poisson random variables, hence using the property of Poisson random variables:

$P\left\{X_1=i-k\right\}=e^{-{\lambda }_1}\frac{{\lambda }_1^{i-k}}{(i-k)!}$

$P\left\{X_3=j-k\right\}=e^{-{\lambda }_3}\frac{{\lambda }_3^{j-k}}{(j-k)!}$

Step12:Final Answer(part c)

Therefore,
$P\{X=i,Y=j\}=\sum _{k=0}^{\min (i,j)}{e}^{-{\lambda }_1}\frac{{\lambda }_1^{i-k}}{(i-k)!}{e}^{-{\lambda }_3}\frac{{\lambda }_3^{j-k}}{(j-k)!}\frac{{\lambda }_2^k}{k!}{e}^{-{\lambda }_2}$