distribution function

Hi people, I am new here and need a help !

Actually this question is taken from home work in the "statistical inference" course, but it's the second week of the course, so the question is still about probability, so I hope it's ok in this forum...

Let X be a random variable with a probability function f:

f(x;alpha)=alpha*e^x*e^(-alpha*e^x)

( yes, it's e^(e^x).... )

A. find the distribution of the random variable Y=exp(X).
B. Let X1,...Xn be a random sample from this distribution.
calculate the distribution of the random variable:
2alpha*sigma(exp(xi))

I failed in both section A and B....
I have passed both "introduction to probability" and "distribution theory", so it's not an easy question, I think.
I hope anyone here can help me, I am trying to solve it for 2 days now !

It's great to find this forum, I will do my best to help other people as well.

is anyone here know a good website with question and answers about statistical inference ? ( advanced inference, what's called statistical theory )

Thanks a lot to all of you that try to help !!

Quote:

---

Originally Posted by WeeG

Let X be a random variable with a probability function f:

f(x;alpha)=alpha*e^x*e^(-alpha*e^x)

( yes, it's e^(e^x).... )

A. find the distribution of the random variable Y=exp(X).

---

Let $\displaystyle X$ be a RV with pdf $\displaystyle p(x)$, and let $\displaystyle g(x)$ be a monotonic increasing function
of $\displaystyle x$. Then the pdf $\displaystyle q(y)$ of RV $\displaystyle Y=g(X)$ is:

$\displaystyle
q(y)=p(g^{-1}(y))\frac{dx}{dy}
$.

This can be quite easily demonstrated by comparing

$\displaystyle \mbox{prob}(y_0<y<y_0+\delta y) \approx q(y_0) \delta y$

with

$\displaystyle \mbox{prob}(x_0<x<x_0+\delta x) \approx p(x_0) \delta x$.

In this case $\displaystyle y=\exp(x)$, so $\displaystyle \frac{dy}{dx}=y$ which maps the real line
onto the positive half line, and

$\displaystyle
f(x,\alpha)=\alpha e^x e^{-\alpha e^x}
$.

So:

$\displaystyle
q(y)=p(\ln(y))/y=\alpha e^{-\alpha y},\ y \ge 0
$,

which is the (negative)exponential distribution.

RonL

Quote:

---

Originally Posted by WeeG

B. Let X1,...Xn be a random sample from this distribution.
calculate the distribution of the random variable:
2alpha*sigma(exp(xi))

---

What we have shown in part A is that Y has an exponential distribution.
So:

$\displaystyle
Z=2 \alpha \sum_1^n e^{X_i}
$

where the $\displaystyle X_i$s are iid RVs with the given distribution, then:

$\displaystyle
Z=2 \alpha \sum_1^n Y_i
$

where the $\displaystyle Y_i$s are iid RV with an exponential distribution, so
the sum has a Gamma distribution with appropriate parameters (standard
result).

RonL