1. ## distribution function F(X)

The lifetime of a product is $Y=5X^{0.7}$ where x has an exponential distribution with mean 1. I need to find the distribution function and pdf of Y.

For the pdf I got $e^{-x}$. Can this be simply integrated to get the distribution function F(Y) to equal $1-e^{-x}$? I obviously need the $Y=5X^{0.7}$ for something, but I'm not sure how to incorporate it besides subbing $x=0.1003393821Y^{10/7}$

Thank you!

2. calculus one
$f_Y(y)=f_X(x)\biggl|{dx\over dy}\biggr|$

3. I'm sorry. I really need a calculus refresher. So you're saying that the pdf of y is equal to the pdf of x multiplied the derivative of pdf of x? like $e^{-x}*(-e^{-x})$?

4. NO....

the pdf of y is equal to the pdf of x multiplied the derivative of THE CHANGE OF VARIABLES

Dependent variables and change of variables
in....
http://en.wikipedia.org/wiki/Probabi...nsity_function

5. thank you for the link. I found this under change of variables. I'm sure I am mistaken, but I though the following means the derivative of the pdf of y equals the derivative of the pdf of x: That would then mean they are equal and both $e^{-x}$, which can't be right. I apologize for being so confused.

6. That's the same as what I wrote.

Use this .... $f_Y(y)=f_X(x)\biggl|{dx\over dy}\biggr|$

Plug in your y for x in $f_X(x)=e^{-x}$, but solve for x first.

Next obtain the derivative ${dx\over dy}$ in terms of y.

Multiply the two and it's over.

7. solving for x in $y=5x^{0.7}$, i got $x=0.100339y^{10/7}$. Plugging that in for the x in $e^{-x}$ and taking the derivative, I got $-0.14334y^{3/7}e^{-0.10034y^{10/7}}$. Multiplying this by the first equation, I ended up with the big messy equation $-0.14334y^{3/7}(e^{-0.10034y^{10/7}})^{2}$

8. so would be my distribution function F(y)?

After differentiating this messy equation I have the p.d.f. f(y)= $-\frac{0.0614(e^{(-0.10034y^{10/7})^{2}}}{y^{4/7}}+0.04109y^{6/7}(e^{(0.10034y^{10/7})^{2}}$.

This all looks too messy to be correct. Can somebody help point me in the right direction? This is confusing me so much. The more I read, the more confused I am getting. Thank you so much for your help!

(cont)
I obviously should have used $x=(\frac{y}{5})^{\frac{10}{7}}$

$f(x)=e^{-(\frac{y}{5})^{\frac{10}{7}}}$
$\frac{dx}{dy}=\frac{-2}{35}*5^{\frac{4}{7}}*y^{\frac{3}{7}}*e^{\frac{-1}{25}*5^{\frac{4}{7}}*y^{\frac{10}{7}}}$
multiplying the two together, I got $\frac{-2}{35}*5^{\frac{4}{7}}*y^{\frac{3}{7}}(e^{\frac{-1}{25}*5^{\frac{5}{7}}*y^{\frac{10}{7}}})^{2}$ which looks a little better but still doesn't seem right. Any help would be extremely appreciated.

9. Originally Posted by laser
The lifetime of a product is $Y=5X^{0.7}$ where x has an exponential distribution with mean 1. I need to find the distribution function and pdf of Y.

For the pdf I got $e^{-x}$. Can this be simply integrated to get the distribution function F(Y) to equal $1-e^{-x}$? I obviously need the $Y=5X^{0.7}$ for something, but I'm not sure how to incorporate it besides subbing $x=0.1003393821Y^{10/7}$

Thank you!