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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 ${位}_i$, for all $i=1,2,3$.

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

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

$={位}_1+{位}_2$

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 $E\left[X\right]={位}_1+{位}_2;E\left[Y\right]={位}_2+{位}_3$

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

Compute, $Cov(X,Y)$

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

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

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

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

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

$Cov(X,Y)={位}_2$

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

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

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:

$P\{X=i,Y=j\}=\underset{k}{鈭�}P\left\{X=i,Y=j鈭�X_2=k\right\}P\left\{X_2=k\right\}$

$=\underset{k}{鈭�}P\left\{X_1=i-k,X_3=j-k鈭�X_2=k\right\}{e}^{-{位}_2}\frac{{位}_2^k}{k!}$

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

$=\underset{k}{鈭�}P\left\{X_1=i-k,X_3=j-k\right\}{e}^{-{位}_2}\frac{{位}_2^k}{k!}$

$=\underset{k}{鈭�}P\left\{X_1=i-k\right\}P\left\{X_3=j-k\right\}{e}^{-{位}_2}\frac{{位}_2^k}{k!}$

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

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

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

Therefore,
$P\{X=i,Y=j\}=\underset{k=0}{\overset{\min (i,j)}{鈭�}}{e}^{-{位}_1}\frac{{位}_1^{i-k}}{(i-k)!}{e}^{-{位}_3}\frac{{位}_3^{j-k}}{(j-k)!}\frac{{位}_2^k}{k!}{e}^{-{位}_2}$