• Apr 27th 2009, 08:36 PM
jennifer1004
Product lifetime is Y=5X^0.7 and X has exponential distribution of $\displaystyle \lambda=1$. Find distribution function and p.d.f. of Y.

My Guess Work:

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

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

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

I really have no idea how to get the right answer to this problem.
• Apr 27th 2009, 11:12 PM
matheagle
this question was asked two days ago
1 y>0, so the absolute value is silly
2 what is R?
3 when you differentiate $\displaystyle (y/5)^{10/7}$ it's better to rewrite this as $\displaystyle {y^{10/7}\over 5^{10/7}}$
so you don't need to use the chain rule here.
• Apr 28th 2009, 02:56 AM
mr fantastic
Quote:

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

My Guess Work:

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

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

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

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