# Exponential distribution

• Mar 21st 2008, 08:16 AM
heathrowjohnny
Exponential distribution
Suppose that $\displaystyle \theta \sim \text{Exp}(\lambda)$. An estimator for $\displaystyle \theta$ is $\displaystyle \bar{X}$. Suppose that we know that $\displaystyle \frac{2n \bar{X}}{\theta} \sim \chi^{2}_{2n}$. Construct a $\displaystyle 100(1-\alpha) \%$ confidence interval for $\displaystyle \theta$.

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

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

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

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

RonL
• Mar 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 $\displaystyle P \left(\chi^{2}_{1- \alpha/2, 2n} < \frac{2n \bar{X}}{\theta} < \chi^{2}_{\alpha/2, 2n} \right) = 1- \alpha$?

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

Originally Posted by heathrowjohnny
Actually aren't they correct?

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

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

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

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

RonL
• Mar 21st 2008, 02:19 PM
heathrowjohnny
Quote:

Originally Posted by CaptainBlack
If you want a $\displaystyle 95\%$ interval $\displaystyle \alpha=5\%$, so

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

RonL

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

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

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

because:

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

RonL
• Mar 22nd 2008, 02:35 PM
heathrowjohnny
So the book probably made a blatant error. They said the first one was $\displaystyle 1- \alpha$.