Question

Suppose the possible values of X are{xi}, the possible values of Y are{yj}, and the possible values of X + Y are {zk}. Let Ak denote the set of all pairs of indices (i,j) suchthatxi + yj =zk;thatis, Ak ={ (i,j) : xi + yj =zk}.

(a) Argue that

$P\left\{X+Y=z_k\right\}=\sum _{(i,j)\in A_k}P\left[X=x_i,Y=y_j\right\}$

(b) Show that

$E[X+Y]=\sum _k\sum _{(i,j)\in A_k}\left(x_i+y_j\right)P\left[X=x_i\right.$

Y =yj}

(c) Using the formula from part (b), argue that

$E[X+Y]=\sum _i\sum _j\left(x_i+y_j\right)P\left\{X=x_i\right.$

Y =yj}

(d) Show that

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

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

(e) Prove that

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

Short answer

a. The probability of the sum of $X$and$Y$is$P\left[X+Y=z_k\right]=\sum _{(i,j)\in A_i} P\left(X=x_i,Y=y_j\right).$

b. The definition of expectation,$E[X+Y]=\sum _k \sum _{(i,j)\in A_k} \left(x_i+y_j\right)P\left\{X=x_i,Y=y_j\right\}$$=\sum _i \sum _{(i,j)\in A_i} \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right).$

c. The formula from part (b), argue that $E[X+Y]=\sum _i \sum _j \left(x_i+y_j\right)P\left\{X=x_i\right.$is $=\sum _i \sum _j \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right).$

d. The probability of $X=\sum _j P\left(X=x_i,Y=y_j\right)$,$Y=\sum _i P\left(X=x_i,Y=y_j\right)$

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

Step-by-step solution

Given Information (Part-a)

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

Solution of the Problem (Part-a)          

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,

$E\left[X\right]=\sum _{x;p\left(x\right)>0} xp\left(x\right)\text{or}E\left[X\right]=\sum _{s\in S} x\left(s\right)p\left(s\right)$

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

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

Prove the Equation (Part-a)

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 $(i,j)\in A_k$ and their sum is

Final Answer(Part-a)

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

$P\left[X+Y=z_k\right]=\sum _{(i,j)\in A_k} P\left(X=x_i,Y=y_j\right)$

Given Information (Part-b)

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

Prove the Equation (Part-b)

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,

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

$=\sum _i z_i\sum _{(i,j)\in A_i} P\left(X=x_i,Y=y_j\right)\left(\text{From part}\right(a\left)\right)$

$=\sum _i \sum _{(i,j)\in {\lambda }_i} \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right)\left(\because z_i=x_i+y_j\right)$

Final Answer (Part-b)

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

Given Information (Part-c)

Given in the question argue that,$E[X+Y]=\sum _i \sum _j \left(x_i+y_j\right)P\left\{X=x_i,\right.Y=yj\}$

Solution of the Problem(Part-c)

From part b,

$E\left[X+Y=z_k\right]=\sum _i \sum _{(i,j)\in A_i} \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right)$

$=\sum _{(0,j)} \sum _i \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right)\text{(rearranging summation)}$

$=\sum _i \sum _j \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right)$

Final Answer (Part-c)

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

Given Information (Part-d)

Given in the question that$P\left(X=x_i\right)=\sum _j P\left(X=x_i,Y=y_j\right)$and

$P\left(Y=y_j\right)=\sum _i P\left\{X=x_i,Y=y_j\right\}$

Find the Probability (Part-d)

The probability of $X$is,

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

$=P\left(X=x_i,\cup \left\{y_j\right\}\right)\left(\because Y\text{takes}\left\{y_j\right\}\right)$

$=\sum _j P\left(X=x_i,Y=y_j\right)$

Similarly, the probability of $Y$is,

$P\left(Y=y_i\right)=P\left(X\in R,Y-y_j\right)(\because \text{There is no condition for}\mathrm{X})$

$=P\left(\cup \left\{x_i\right\},Y=y_j,\right)\left(\because Y\text{takes}\left\{y_j\right\}\right)$

$=\sum _i P\left(X=x_i,Y=y_j\right)$

Final Answer (Part-d)

The probability of $X$$=\sum _j P\left(X=x_i,Y=y_j\right)$

the probability of$Y$$=\sum _i P\left(X=x_i,Y=y_j\right)$

Given Information (Part-e)

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

Prove the Equation (Part-e)

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

$E[X+Y]=\sum _i \sum _j \left(x_i+y_j\right)P\left(X=x_i,Y=y_j\right)$

$=\sum _i \sum _j \left(x_i\right)P\left(X=x_i,Y=y_j\right)+\sum _i \sum _j \left(y_j\right)P\left(X=x_i,Y=y_j\right)$

$=\sum _i \left(x_i\right)\sum _j P\left(X=x_i,Y=y_j\right)+\sum _j \left(y_j\right)\sum _i P\left(X=x_i,Y=y_j\right)$

$=\sum _i \left(x_i\right)P\left(X=x_i\right)+\sum _j \left(y_j\right)P\left(Y=y_j\right)$

We get,

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

Final Answer (Part-e)

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