Independent solution

How to solve this Mixed Distributions question

Setup

Setup

The law of X contains both a point mass 1/2 at x=1 and a continuous density x-1 on (1,2). Both components enter each moment.

P(X=1)=12P(X=1)=\frac12
fX(x)=x−1,1<x<2f_X(x)=x-1,\qquad 1<x<2

Model

Model

Compute the first and second raw moments by adding the atom's contribution to the corresponding integral over the continuous part.

E[X]=12+∫12x(x−1) dxE[X]=\frac12+\int_1^2x(x-1)\,dx
E[X2]=12+∫12x2(x−1) dxE[X^2]=\frac12+\int_1^2x^2(x-1)\,dx

Compute

Compute

The first moment is 4/3 and the second raw moment is 23/12. Subtracting the squared mean gives variance 5/36.

E[X]=43E[X]=\frac43
E[X2]=2312E[X^2]=\frac{23}{12}
Var⁡(X)=2312−(43)2=536\operatorname{Var}(X)=\frac{23}{12}-\left(\frac43\right)^2=\frac5{36}

Answer

Answer

Therefore Var(X)=5/36, which is choice C.

536(C)\boxed{\frac5{36}\quad\text{(C)}}