91Ó°ÊÓ

Given in the question that$P\left\{X+Y=z_k\right\}=\underset{(i,j)∈A_k}{∑} P\left\{X=x_i,Y=y_j\right\}$

If $X$is a discrete random variable with probability mass function $p\left(x\right),$then the expectation, or the expected value, of $X$is defined by,

$E\left[X\right]=\underset{x;p\left(x\right)>0}{∑} xp\left(x\right)\text{or}E\left[X\right]=\underset{s∈S}{∑} x\left(s\right)p\left(s\right)$

If $X,Y$are any two discrete random variables then

$E[X+Y]=E\left(X\right)+E\left(Y\right)$

Here the possible values of the variable $X$are $\left\{x_i\right\}$and the possible values of $\left\{y_i\right\}$and the possible values of $X+Y$are $\left\{z_k\right\}$

Let $A_k$denote the set of all pairs of indices $(i,j)$such that $x_i+y_i=z_k$; that is,

$A_k=\left\{(i,j):x_i+y_j=z_k\right\}$

Therefore $X+Y$ takes values independently, in such a way that the pair of indices $(i,j)∈A_k$ and their sum is

Therefore, the probability of the sum of $X$ and $Y$ is,

Given in the question that$E[X+Y]=\underset{k}{∑} \underset{(i,j)∈A_k}{∑} \left(x_i+y_j\right)P\left\{X=x_{i,}\left.Y=y_j\right\}\right.$$E[X+Y]=\underset{k}{∑} \underset{(i,j)∈A_k}{∑} \left(x_i+y_j\right)P\left\{X=x_{i,}\left.Y=y_j\right\}\right.$

Here the possible values of the variable $X$are $\left\{x_i\right\}$and the possible values of $\left\{y_i\right\}$and the possible values of $X+Y$are$\left\{z_k\right\}.$Let'srole="math" $A_k$denote the set of all pairs of indices $(i,j)$such that $x_i+y_i=z_k$;

That is,$A_k=\left\{(i,j):x_i+y_j=z_k\right\}$

Therefore, using the definition of expectation,

$E\left[X+Y=z_k\right]=∑z_{k}P\left(X+Y=z_k\right)$

$=\underset{i}{∑} z_i\underset{(i,j)∈A_i}{∑} P\left(X=x_i,Y=y_j\right)\left(\text{From part}\right(a\left)\right)$

$=\underset{i}{∑} \underset{(i,j)∈{\mathrm{λ}}_i}{∑} \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right)\left(âˆ\mu z_i=x_i+y_j\right)$

The definition of expectation,$E[X+Y]=\underset{k}{∑} \underset{(i,j)∈A_k}{∑} \left(x_i+y_j\right)P\left\{X=x_i,Y=y_j\right\}$$=\underset{i}{∑} \underset{(,j)k{x}_i}{∑} \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right).$

Given in the question argue that,$E[X+Y]=\underset{i}{∑} \underset{j}{∑} \left(x_i+y_j\right)P\left\{X=x_i,\right.Y=yj\}$

From part b,

$E\left[X+Y=z_k\right]=\underset{i}{∑} \underset{(i,j)∈A_i}{∑} \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right)$

$=\underset{(0,j)}{∑} \underset{i}{∑} \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right)\text{(rearranging summation)}$

$=\underset{i}{∑} \underset{j}{∑} \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right)$

The formula from part (b), argue that $E[X+Y]=\underset{i}{∑} \underset{j}{∑} \left(x_i+y_j\right)P\left\{X=x_i,Y=yj\right\}$is$=\underset{i}{∑} \underset{j}{∑} \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right).$

Given in the question that$P\left(X=x_i\right)=\underset{j}{∑} P\left(X=x_i,Y=y_j\right)$and

$P\left(Y=y_j\right)=\underset{i}{∑} P\left\{X=x_i,Y=y_j\right\}$

The probability of $X$is,

$P\left(X=x_i\right)=P\left(X=x_i,Y∈R\right)(âˆ\mu \text{There is no condition for}Y)$

$=P\left(X=x_i,∪\left\{y_j\right\}\right)\left(âˆ\mu Y\text{takes}\left\{y_j\right\}\right)$

$=\underset{j}{∑} P\left(X=x_i,Y=y_j\right)$

Similarly, the probability of $Y$is,

$P\left(Y=y_i\right)=P\left(X∈R,Y−y_j\right)(âˆ\mu \text{There is no condition for}\mathrm{X})$

$=P\left(∪\left\{x_i\right\},Y=y_j,\right)\left(âˆ\mu Y\text{takes}\left\{y_j\right\}\right)$

$=\underset{i}{∑} P\left(X=x_i,Y=y_j\right)$

The probability of $X$$=\underset{j}{∑} P\left(X=x_i,Y=y_j\right)$

the probability of$Y$$=\underset{i}{∑} P\left(X=x_i,Y=y_j\right)$

Given in the question, we prove that$E[X+Y]=E\left[X\right]+E\left[Y\right]$

Using the definitions of$X$and $Y$, the sum of expectations of $X$and $Y$is,

$E[X+Y]=\underset{i}{∑} \underset{j}{∑} \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right)$

$=\underset{i}{∑} \underset{j}{∑} \left(x_i\right)P\left(X=x_i,Y=y_j\right)+\underset{i}{∑} \underset{j}{∑} \left(y_j\right)P\left(X=x_i,Y=y_j\right)$

$=\underset{i}{∑} \left(x_i\right)\underset{j}{∑} P\left(X=x_i,Y=y_j\right)+\underset{j}{∑} \left(y_j\right)\underset{i}{∑} P\left(X=x_i,Y=y_j\right)$

$=\underset{i}{∑} \left(x_i\right)P\left(X=x_i\right)+\underset{j}{∑} \left(y_j\right)P\left(Y=y_j\right)$

We get,

$=E\left(X\right)+E\left(Y\right)$

Using the definitions of $X$and $Y,$the sum of expectations of$X$and$Y$, We proved $E[X+Y]=E\left(X\right)+E\left(Y\right)$.