Variance of First Order Statistic of Exponential Distribution

Question

If $X_1$ and $X_2$ constitute a random sample of size $n=2$ from an exponential population, find the efficiency of $2Y_1$ relative to $\bar{X}$, where $Y_1$ is the first order statistic and $2Y_1$ and $\bar{X}$ are both unbiased estimators of the parameter $\theta$.

I keep going in circles on this problem.What is really giving me trouble is finding the variance of $Y_1$. I know the CDF for an exponential distribution will be $F(x)=1-e^{-\lambda x}$, in this particular case $\lambda=\frac{1}{\theta}$, so the pdf for the first order statistic $g(y_1)=2(1-F(y_1))=2e^{-\lambda y_1}$, right? Going from there, finding the variance of $2Y_1$ in order to find the efficiency should be as follow, but I'm really struggling with the calculations:

$$V(2Y_1)=4V(Y_1)=4[E(Y_1^2)-E(Y_1)^2]$$

Once I have that, I know the final relative efficiency will be $\frac{V(\bar{X})}{V(2Y_1)}$ and that $V(\bar{X})=\frac{\theta^2}{2}$. I just need a little help getting in the right direction here; I feel so off.

Answer 1

The pdf of $Y_1$ is $2\lambda e^{-2\lambda x}, \ x\geq 0.$ (And not $2e^{-\lambda x}$.)

Let's prove this statement first.

By definition, $$Y_1=\min(X_1,X_2).$$

So,

$$F_{Y_1}(x)=P(X_1

First

$$P(X_1

Because of symmetry reasons the second component of $(1)$ is the same. So, we have

$$F_{Y_1}(x)=1-e^{-2\lambda x}.$$

From here the pdf is

$$f_{X_1}(x)=2\lambda e^{-2\lambda x}.$$

This is an exponential distribution with parameter $2\lambda$. The expectation and the variance are $\frac1{2\lambda}$ and $\frac1{4\lambda^2}$, respectively.

Finally, the variance of $2Y_1$ is $$\frac1{\lambda^2}.$$

Answer 2

For $y\ge 0,$ $$ \Pr(Y_1>y) = \Pr(X_1>y\ \&\ X_2>y) = (\Pr(X_1>y))^2 = (e^{-\lambda y} )^2 = e^{-2\lambda y}. $$ Therefore $$ f_{Y_1}(y) = \frac d {dy} \Pr(Y_1\le y) = \frac d {dy} (1-e^{-2\lambda y}) = 2\lambda e^{-2\lambda y} \quad \text{ for } y>0. $$ And $\operatorname{var}(2Y_1) = 4\operatorname{var}(Y_2).$