Independent solution

How to solve this Marginal Distributions question

Setup

Setup

Marginalize the supplied joint mass function over every allowed value of the second coordinate.

pX(x)=y=0xx+2y40,x=0,1,2,3p_X(x)=\sum_{y=0}^{x}\frac{x+2y}{40},\qquad x=0,1,2,3

Model

Model

The finite arithmetic sum reduces to a simple marginal mass function.

y=0x(x+2y)=x(x+1)+x(x+1)=2x(x+1)\sum_{y=0}^{x}(x+2y)=x(x+1)+x(x+1)=2x(x+1)
pX(x)=x(x+1)20p_X(x)=\frac{x(x+1)}{20}

Compute

Compute

Calculate both raw moments from the marginal probabilities.

Pr(X=1)=0.1,Pr(X=2)=0.3,Pr(X=3)=0.6\Pr(X=1)=0.1,\qquad \Pr(X=2)=0.3,\qquad \Pr(X=3)=0.6
E[X]=1(0.1)+2(0.3)+3(0.6)=2.5\operatorname{E}[X]=1(0.1)+2(0.3)+3(0.6)=2.5
E[X2]=12(0.1)+22(0.3)+32(0.6)=6.7\operatorname{E}[X^2]=1^2(0.1)+2^2(0.3)+3^2(0.6)=6.7
Var(X)=6.7(2.5)2=0.45\operatorname{Var}(X)=6.7-(2.5)^2=0.45

Answer

Answer

The marginal variance is 0.45.

Var(X)=0.45(A)\boxed{\operatorname{Var}(X)=0.45\quad\text{(A)}}