0
$\begingroup$

Edit: it was a typo. I correted the question and will put my solution to it.

I need to find a confindence interval for a given parameter. Let $X_1 \ldots X_n$ be i.i.d rv.'s with a p.d.f. given by:

$$ f(x|\theta) = \frac{1}{2\theta} e^{-|x|/\theta}$$

where the parameter $\theta \in (0, \infty)$.

Show that $$ 2 \frac{\sum_{j=1}^{n} |X_j|}{\theta} \sim \chi^2_{2n} $$ has a chi-squared distribution so it can be used as a pivot quantity, and create a confidence interval from that.

$\endgroup$
0
$\begingroup$

Let $Z = |X|$. Then:

$$\mathbb{P}(Z\leq t) = \mathbb{P}(-t \leq X \leq t) = \int_{-t}^{t} \frac{1}{2\theta}e^{|x|/\theta}dx = 1-e^{-t/\theta} $$

So the probability density function will be given by:

$$ f_Z(z) = \frac{1}{\theta} e^{-z/\theta} \ \mathbb{I}_{(0,\infty)}(z) $$

From that, we can evaluate the MGF:

$$ \phi_Z(t) = \mathbb{E}(e^{tZ}) = \int_{0}^{\infty} \frac{1}{\theta }e^{tz-z/\theta}dz = \frac{1}{1-t\theta} $$

Which results:

$$ \phi_{\frac{2\sum_{j=1}^{n} Z_j}{\theta}}(t) = \phi_{\sum_{j=1}^{n} Z_j}\left(\frac{2t}{\theta}\right) = \left(1-2t\right)^{-n}$$

And that's precisely the MGF of a $ \chi_{2n}^2 $ distribution, and is a pivot quantity. It can then be used to create a confidence interval for the parameter $\theta$, meaning:

If we choose $a,b \in \mathbb{R}$ such that:

$$ \mathbb{P}(a \leq \chi^{2}_{2n} \leq b) = 1-\alpha $$

then:

$$ \left[\frac{2\sum_{j=1}^{n} |X_j|}{b}; \frac{2\sum_{j=1}^{n} |X_j|}{a} \right] $$

is a confidence interval for the parameter $\theta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.