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...
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....
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}$
which is what you had
[10/7*y^(3/7)*e^(-y/5)^10/7] / 5^(10/7)
Popular question. Asked and answered (in different ways) in the following threads:
http://www.mathhelpforum.com/math-he...ion-p-d-f.html
http://www.mathhelpforum.com/math-he...ction-f-x.html
http://www.mathhelpforum.com/math-he...-lifetime.html