# Thread: Weibull & Exponential distribution

1. ## Weibull & Exponential distribution

$X$ is a random variable with weibull distribution given with

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

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

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

2. you have the right approach

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

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.

3. Originally Posted by losm1
$X$ is a random variable with weibull distribution given with

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

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

I believe that $G(Y)=P(Y \leq y)$ should be expressed in terms of $X$ , 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:

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

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

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.

4. Originally Posted by SpringFan25
you have the right approach

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

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 $\lambda=0.5$

I hope that it's correct.