# Weibull & Exponential distribution

• Jun 19th 2010, 02:19 AM
losm1
Weibull & Exponential distribution
$\displaystyle X$ is a random variable with weibull distribution given with

$\displaystyle f(x;\alpha ,\beta) = \begin{cases} \frac{\alpha}{\beta^\alpha}x^{\alpha-1}e^{-(x/\beta)^{\alpha}} & x>0\\ 0 & x=0\end{cases}$

Let $\displaystyle \alpha=2$ and $\displaystyle Y=2X^2/\beta^2$.

Show that $\displaystyle Y$ has exponential distribution and find value of parameter $\displaystyle \lambda$.

I believe that $\displaystyle G(Y)=P(Y \leq y)$ should be expressed in terms of $\displaystyle X$ , but I don't know where to start.
• Jun 19th 2010, 03:29 AM
SpringFan25
you have the right approach

Lets find P(Y<y)
$\displaystyle P(Y \leq y) = P(2x^2/B^2 \leq y) = P(x \leq \sqrt {0.5 B^2 y}) = F(\sqrt {0.5 B^2 y})$

Find (or look up) F(x) then subsititute these values in to get the CDF of Y. Hopefully it will look like an exponential distribution.
• Jun 19th 2010, 03:30 AM
mr fantastic
Quote:

Originally Posted by losm1
$\displaystyle X$ is a random variable with weibull distribution given with

$\displaystyle f(x;\alpha ,\beta) = \begin{cases} \frac{\alpha}{\beta^\alpha}x^{\alpha-1}e^{-(x/\beta)^{\alpha}} & x>0\\ 0 & x=0\end{cases}$

Let $\displaystyle \alpha=2$ and $\displaystyle Y=2X^2/\beta^2$.

Show that $\displaystyle Y$ has exponential distribution and find value of parameter $\displaystyle \lambda$.

I believe that $\displaystyle G(Y)=P(Y \leq y)$ should be expressed in terms of $\displaystyle X$ , but I don't know where to start.

Since you know in advance that the answer is going to be a well-known distribution, a simple approach would be to calculate the moment generating function of Y ....

But if you're determined to use a distribution function approach, then you should start by noting that the cdf of Y is:

$\displaystyle F(y) = \Pr(Y \leq y) = \Pr \left( \frac{2 X^2}{\beta^2} \leq y \right) = \Pr \left( - \frac{\beta}{\sqrt{2}} \sqrt{y} \leq X \leq \frac{\beta}{\sqrt{2}} \sqrt{y} \right)$

$\displaystyle = \Pr \left( 0 < X \leq \frac{\beta}{\sqrt{2}} \sqrt{y} \right)$ since the support of X is $\displaystyle x \geq 0$.

Now set up the required integral (no need to integrate by the way ....) and then differentiate using the chain rule and the Fundamental Theorem of Calculus.
• Jun 19th 2010, 04:14 AM
losm1
Quote:

Originally Posted by SpringFan25
you have the right approach

Lets find P(Y<y)
$\displaystyle P(Y \leq y) = P(2x^2/B^2 \leq y) = P(x \leq \sqrt {0.5 B^2 y}) = F(\sqrt {0.5 B^2 y})$

Find (or look up) F(x) then subsititute these values in to get the CDF of Y. Hopefully it will look like an exponential distribution.

Thanks, I just plugged in your result in weibull CDF and got exponential CDF with $\displaystyle \lambda=0.5$

I hope that it's correct.