Exponential distribution

• March 21st 2008, 09:16 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, 12:03 PM
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, 03:09 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, 03:10 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, 03:19 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, 01:53 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, 03:35 PM
heathrowjohnny
So the book probably made a blatant error. They said the first one was $1- \alpha$.