1. ## product lifetime

Product lifetime is Y=5X^0.7 and X has exponential distribution of $\lambda=1$. Find distribution function and p.d.f. of Y.

My Guess Work:

So far, I know $x=(y/5)^{10/7}$ and $f(x)=e^{-x}$ since the mean is 1.

When I take the derivative of the equation after substituting $(y/5)^{10/7}$ for x, I get $(-2/35)5^{4/7}y^{3/7}e^{(-1/25)5^{4/7}y^{10/7}}$.

Then multiplying the absolute value of this equation by the original equation of $e^{(y/5)^{(10/7)}}$, I get $(2/25)e^{(-1/25)(5^{4/7}y^{10/7}}5^{4/5}e^{(-1/25)5^{4/7}R(y^{10/7})}|y|^{3/7}$

I really have no idea how to get the right answer to this problem.

2. this question was asked two days ago
1 y>0, so the absolute value is silly
2 what is R?
3 when you differentiate $(y/5)^{10/7}$ it's better to rewrite this as ${y^{10/7}\over 5^{10/7}}$
so you don't need to use the chain rule here.

3. Originally Posted by jennifer1004
Product lifetime is Y=5X^0.7 and X has exponential distribution of $\lambda=1$. Find distribution function and p.d.f. of Y.

My Guess Work:

So far, I know $x=(y/5)^{10/7}$ and $f(x)=e^{-x}$ since the mean is 1.

When I take the derivative of the equation after substituting $(y/5)^{10/7}$ for x, I get $(-2/35)5^{4/7}y^{3/7}e^{(-1/25)5^{4/7}y^{10/7}}$.

Then multiplying the absolute value of this equation by the original equation of $e^{(y/5)^{(10/7)}}$, I get $(2/25)e^{(-1/25)(5^{4/7}y^{10/7}}5^{4/5}e^{(-1/25)5^{4/7}R(y^{10/7})}|y|^{3/7}$

I really have no idea how to get the right answer to this problem.
Popular question. Asked and answered (in different ways) in the following threads:

http://www.mathhelpforum.com/math-he...tial-dist.html

http://www.mathhelpforum.com/math-he...ion-p-d-f.html

http://www.mathhelpforum.com/math-he...ction-f-x.html