Exponential distribution

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March 21st 2008, 08:16 AM
heathrowjohnny

Exponential distribution

Suppose that $\theta \sim \text{Exp}(\lambda)$ . An estimator for $\theta$ is $\bar{X}$ . Suppose that we know that $\frac{2n \bar{X}}{\theta} \sim \chi^{2}_{2n}$ . Construct a $100(1-\alpha) \%$ confidence interval for $\theta$ .

Would it be $\frac{2n \bar{X}}{\chi^{2}_{\alpha/2, 2n}} < \theta < \frac{2n \bar{X}}{\chi^{2}_{1-\alpha/2, 2n}}$ ?
March 21st 2008, 11:03 AM
CaptainBlack

Quote:

Originally Posted by heathrowjohnny

Suppose that $\theta \sim \text{Exp}(\lambda)$ . An estimator for $\theta$ is $\bar{X}$ . Suppose that we know that $\frac{2n \bar{X}}{\theta} \sim \chi^{2}_{2n}$ . Construct a $100(1-\alpha) \%$ confidence interval for $\theta$ .

Would it be $\frac{2n \bar{X}}{\chi^{2}_{\alpha/2, 2n}} < \theta < \frac{2n \bar{X}}{\chi^{2}_{1-\alpha/2, 2n}}$ ?

Now I think that those inequalities are the wrong way around.

RonL
March 21st 2008, 02:09 PM
heathrowjohnny

Quote:

Originally Posted by CaptainBlack

Now I think that those inequalities are the wrong way around.

RonL

Actually aren't they correct?

Wouldn't it be $P \left(\chi^{2}_{1- \alpha/2, 2n} < \frac{2n \bar{X}}{\theta} < \chi^{2}_{\alpha/2, 2n} \right) = 1- \alpha$ ?

And this is equivalent to what I have written? Or maybe I made a mistake.
March 21st 2008, 02:10 PM
CaptainBlack

Quote:

Originally Posted by heathrowjohnny

Actually aren't they correct?

Wouldn't it be $P \left(\chi^{2}_{1- \alpha/2, 2n} < \frac{2n \bar{X}}{\theta} < \chi^{2}_{\alpha/2, 2n} \right) = 1- \alpha$ ?

And this is equivalent to what I have written? Or maybe I made a mistake.

If you want a $95\%$ interval $\alpha=5\%$ , so

$\chi^{2}_{1- \alpha/2, 2n}=\chi^2_{97.5\%,2n} > \chi^{2}_{2.5\%, 2n} =\chi^{2}_{\alpha/2, 2n}$ ?

RonL
March 21st 2008, 02:19 PM
heathrowjohnny

Quote:

Originally Posted by CaptainBlack

If you want a $95\%$ interval $\alpha=5\%$ , so

$\chi^{2}_{1- \alpha/2, 2n}=\chi^2_{97.5\%,2n} > \chi^{2}_{2.5\%, 2n} =\chi^{2}_{\alpha/2, 2n}$ ?

RonL

But now that we are dividing those terms, the sign changes? Isn't it similar to the following: $P \left(\chi^{2}_{1- \alpha/2, n-1} < \frac{(n-1)S^{2}}{\sigma^{2}} < \chi^{2}_{\alpha/2, n-1} \right) = 1- \alpha$ , and so $\frac{(n-1)S^{2}}{\chi^{2}_{\alpha/2, n-1}} < \sigma^{2} < \frac{(n-1)S^{2}}{\chi^{2}_{1- \alpha/2, n-1}}$ ?
March 22nd 2008, 12:53 AM
CaptainBlack

Quote:

Originally Posted by heathrowjohnny

But now that we are dividing those terms, the sign changes? Isn't it similar to the following: $P \left(\chi^{2}_{1- \alpha/2, n-1} < \frac{(n-1)S^{2}}{\sigma^{2}} < \chi^{2}_{\alpha/2, n-1} \right) = 1- \alpha$ , and so $\frac{(n-1)S^{2}}{\chi^{2}_{\alpha/2, n-1}} < \sigma^{2} < \frac{(n-1)S^{2}}{\chi^{2}_{1- \alpha/2, n-1}}$ ?

$P \left(\chi^{2}_{97.5\%, n-1} < \frac{(n-1)S^{2}}{\sigma^{2}} < \chi^{2}_{2.5\%, n-1} \right) = 0$

$P \left(\chi^{2}_{2.5\%, n-1} < \frac{(n-1)S^{2}}{\sigma^{2}} < \chi^{2}_{97.5\%, n-1} \right) \ne 0$

because:

$\chi^{2}_{2.5\%, n-1} < \chi^{2}_{97.5\%, n-1}$

RonL
March 22nd 2008, 02:35 PM
heathrowjohnny

So the book probably made a blatant error. They said the first one was $1- \alpha$ .