# Thread: The expected value of this estimator, involving the minimum statistic of an exp dstrn

1. ## The expected value of this estimator, involving the minimum statistic of an exp dstrn

Suppose that Y1, Y2, .... Yn ~exp(B). Define two estimators for B as:

B1 = nY(1) ; B2 = 1/n Summation from i-n of Y, or Y-bar.

I'm trying to determine that the first estimator is unbiased, but I've returned expected values of it as n3B.

Y(1) =n[1 -Fy]n-1fy,
= n (e-y/B)n-1(1/B)(e-y/B)
= n/B (e-y/B)n
= (n2/B)(e-y/B)

E[B1] = E[n* (n2/B)(e-y/B)] = n3E[(1/B)(e-y/B)] = n3B.

Obviously this is incorrect. Where am I misunderstanding something? I've gone over the minimum order statistic function like ten times and the book we are using just glosses over the expectation of it.

Edit: I've noticed at least one mistake, putting (e^(-y/B))^n as n(e^(-y/B)) is not correct algebra. but, I'm still stuck.

2. ## Re: The expected value of this estimator, involving the minimum statistic of an exp d

Note that I am assuming that the $Y_i$s are independent!!

First, find the CDF of $Y_{(1)}$:

\begin{align*}F_{Y_{(1)}}(y) &= P(Y_{(1)}\leq y) \\ &= 1 - P(Y_{(1)}> y) \\ &= 1-P\left([Y_1>y] \cap [Y_2>y] \cap \ldots \cap [Y_n>y]\right) \\ &= 1-\left[P(Y_1>y)\cdot P(Y_2>y)\cdots P(Y_n>y)\right] \\ &= 1 - \left[(e^{-\tfrac{y}{\beta}}) \cdot (e^{-\tfrac{y}{\beta}}) \cdots (e^{-\tfrac{y}{\beta}})\right] \\ &= 1 - e^{-\left(\tfrac{n}{\beta}\right)y} \end{align*}

The derivative of the CDF with respect to $y$ gives us the PDF:

\begin{align*}f_{Y_{(1)}}(y) &= \frac{n}{\beta} \cdot e^{-\left(\tfrac{n}{\beta}\right)y} \end{align*}

We now compute $E[Y_{(1)}]$ using integration by parts.

For neatness, we use $\lambda=\tfrac{n}{\beta}$ for the integral. We will use $u=y, u^\prime=1, v=-e^{-\lambda y}, v^\prime=\lambda e^{-\lambda y}$.

\begin{align*}E[Y_{(1)}] &= \int^\infty_0 y\cdot f_{Y_{(1)}}(y) dy \\ &= \int^\infty_0 y \lambda e^{-\lambda y} dy \\ &= \left[-ye^{-\lambda y} + \int e^{-\lambda y} dy \right]^{\infty}_{0} \\ &= \frac{1}{\lambda}\\ &= \frac{1}{n/\beta}\\ &= \frac{\beta}{n} \end{align*}

Got it from here?

Best,
Andy