0

All random variables here are iid. We have that $$f(x;\theta) = \alpha \mbox{ exp}\left[ -\alpha (x - \beta) \right] \times \mathbb{I} \left\{x \geq \beta \right\} $$

for $\alpha > 0$ and $\beta \in \mathbb{R}$.

I have found maximum likelihood estimators for both.

$$\hat{\alpha} = (\bar{{\bf x}} - \hat{\beta})$$ $$\hat{\beta} = \mbox{min}({\bf x})$$ where bold x is the vector of observations of X.

I need to show the sampling distribution of $\hat{\beta} - \beta$ is

$$\mathbb{P}(\hat{\beta} - \beta \geq v) = \mbox{exp}(-\alpha n \times v \mathbb{I} \left\{ 0 \leq v\right\})$$

Any help would be appreciated. I have attempted to show

$$\mathbb{P}(\hat{\beta} - \beta \geq v) = \mathbb{P}(\mbox{min}({\bf x}) - \beta \geq v) = \prod^n_{i = 1}\mathbb{P}(x_i - \beta \geq v)$$

And go from here, but without much success.

Thanks.

1 Answers1

0

The cumulative distribution function and the probability density function of the minimum from an i.i.d sample of size $n$ are rather well-known:

$$F_{\hat \beta}(\hat \beta) = 1- \left(1-F_X(\hat \beta)\right)^n$$ and

$$f_{\hat \beta}(\hat \beta) = n\left(1-F_X(\hat \beta)\right)^{n-1}f_X(\hat \beta)$$

You know $f_X(x)$ and it is easy to integrate it so that you can obtain also $F_X(x)$. Can you take it from here?

ADDENDUM

Now that the OP has found the way I am completing this.

We have $$F_X(x) = \int_{-\infty}^{x}\alpha \mbox{ exp}\left[ -\alpha (t - \beta) \right] \cdot \mathbb{I} \left\{t \geq \beta \right\} dt$$

$$=\int_{\beta}^{x}\alpha \mbox{ exp}\left[ -\alpha (t - \beta) \right] dt = -\mbox{ exp}\left[ -\alpha (t - \beta) \right]\Big |_{\beta}^x = 1-\mbox{ exp}\left[ -\alpha (x - \beta) \right]$$

So

$$F_{\hat \beta}(\hat \beta) = 1- \left(\mbox{ exp}\left[ -\alpha (\hat \beta - \beta) \right]\right)^n$$

and $$f_{\hat \beta}(\hat \beta) = n\left(\mbox{ exp}\left[ -\alpha (\hat \beta - \beta) \right]\right)^{n-1}\cdot \alpha \mbox{ exp}\left[ -\alpha (\hat \beta - \beta) \right] \cdot \mathbb{I} \left\{\hat \beta \geq \beta \right\}$$

$$=\alpha n\left(\mbox{ exp}\left[ -\alpha n(\hat \beta - \beta) \right]\right)\cdot \mathbb{I} \left\{\hat \beta \geq \beta \right\}$$

Setting $v\equiv \hat \beta - \beta$ we obtain

$$f_{V}(v) = \alpha n\left(\mbox{ exp}\left[ -\alpha nv \right]\right)\cdot \mathbb{I} \left\{v\ge 0 \right\}$$

Then

$$P(\hat \beta - \beta\ge v) = P(V\ge v) = 1- F_{V}(v) = 1-\int_{-\infty}^v\alpha n\left(\mbox{ exp}\left[ -\alpha nt \right]\right)\cdot \mathbb{I} \left\{t\ge 0 \right\}dt$$ $$=1-\int_{0}^v\alpha n\left(\mbox{ exp}\left[ -\alpha nt \right]\right)dt = 1 +\mbox{ exp}\left[ -\alpha nt\right]\Big|_0^v = \mbox{ exp}\left[ -\alpha nv\right], \;; v\ge 0$$ or

$$P(\hat \beta - \beta\ge v) = \mbox{ exp}\left[ -\alpha nv\cdot\mathbb{I} \left\{v\ge 0 \right\}\right]$$

Alecos Papadopoulos
  • 10,388