1. ## Another exponential dist.

The lifetime of a product is Y = 5X^0.7 , where X has an exponential dist. with mean = 1

Find the distribution function and pdf of Y

I'm not even sure where to start here...

2. The lifetime of a product is Y = 5X^0.7 , where X has an exponential dist. with mean = 1

So the cdf, $\displaystyle F(x) = \mathbf{P}(Y\leq x) = \mathbf{P}(5X^{0.7}\leq x) = \mathbf{P}(X\leq (x/5)^{1/.7})$

The last expression is in the form of the CDF of an exponential. ($\displaystyle \mathbf{P}(X\leq x_o) = 1 - exp(-x/\mu)$) And so

$\displaystyle = 1-exp(-(x/5)^{1/.7})$

To get the pdf, just take the derivative of the CDF....

3. I did that, but it's not the correct answer.
Unless I'm doing something wrong

[10/7*y^(3/7)*e^(-y/5)^10/7] / 5^(10/7)

Looks like your answer is pretty close, so I assume it's me doing something wrong. Thanks for the help

4. derivative of the CDF:
$\displaystyle \frac{d}{dx}\left( 1-\exp(-(x/5)^{1/.7}) \right)$

derivative of the sum is sum of the derivatives

derivative of constant is 0

1/.7 = 10/7

$\displaystyle = 0 +\frac{d}{dx}\left(-\exp(-(x/5)^{10/7}) \right)$

Now $\displaystyle \frac{d}{dx}\exp(u) = exp(u)\frac{du}{dx}$ (derivative of exp with chain rule). So

$\displaystyle = -\exp(-(x/5)^{10/7}) \frac{d}{dx} $$\displaystyle \left(-(x/5)^{10/7}\right) Then \displaystyle = -\exp(-(x/5)^{10/7})$$\displaystyle \frac{-10}{7}(x/5)^{10/7-1}\frac{d}{dx}(x/5)$

$\displaystyle = $$\displaystyle \frac{10}{7} exp(-(x/5)^{10/7}) x^{3/7}\frac{1}{5^{3/7}5} \displaystyle =$$\displaystyle \frac{10}{7} \frac{1}{5^{10/7}}exp(-(x/5)^{10/7}) x^{3/7}$

[10/7*y^(3/7)*e^(-y/5)^10/7] / 5^(10/7)

5. Yea, I actually figured it out like an hour ago. Been too many years since calculus; thanks for all your help!

6. Originally Posted by amor_vincit_omnia
The lifetime of a product is Y = 5X^0.7 , where X has an exponential dist. with mean = 1

Find the distribution function and pdf of Y

I'm not even sure where to start here...