Independent solution

How to solve this Exponential Distribution question

Answer in brief

The second moment condition first forces the mean of Y to equal 2. Substitution into the remaining mean-variance identity gives alpha squared minus alpha minus 2 equal to zero, whose positive root is 2, so the answer is D.

Setup

Setup

Use the fact that an exponential variable with mean m has variance m squared.

E[U]=α+β+4E[U]=\alpha+\beta+4
Var⁡(U)=α2+β2+16\operatorname{Var}(U)=\alpha^2+\beta^2+16
E[V]=α−β,Var⁡(V)=α2+β2E[V]=\alpha-\beta,\qquad \operatorname{Var}(V)=\alpha^2+\beta^2

Model

Model

The relation involving the differences of the two means and variances isolates beta.

E[U]−E[V]=2β+4E[U]-E[V]=2\beta+4
Var⁡(U)−Var⁡(V)2=8\frac{\operatorname{Var}(U)-\operatorname{Var}(V)}{2}=8
2β+4=82\beta+4=8

Compute

Compute

Solve for beta, then apply the remaining equality and retain the positive exponential mean.

β=2\beta=2
α+6=α2+4\alpha+6=\alpha^2+4
α2−α−2=0\alpha^2-\alpha-2=0
α∈{2,−1}\alpha\in\{2,-1\}

Answer

Answer

An exponential mean must be positive, so alpha equals 2.

α=2.0(D)\boxed{\alpha=2.0\quad\text{(D)}}