# Probability Density Function

• May 29th 2008, 09:20 PM
bloomsgal8
Probability Density Function
I'm working on a practice midterm and this question has me a bit stumped. Can anyone please help in working it out? Thank you!!

f(x) = ae^(-ax) with 0<equalto x < infinity

where a is constant, show that the mean and standard deviation are both 1/a

Thank you.
• May 29th 2008, 10:00 PM
mr fantastic
Quote:

Originally Posted by bloomsgal8
I'm working on a practice midterm and this question has me a bit stumped. Can anyone please help in working it out? Thank you!!

f(x) = ae^(-ax) with 0<equalto x < infinity

where a is constant, show that the mean and standard deviation are both 1/a

Thank you.

Mean: $\displaystyle \mu = a \int_0^{+\infty} x \, e^{-ax} \, dx$.

Do this integral using integration by parts.

Variance: $\displaystyle E(X^2) = a \int_0^{+\infty} x^2 \, e^{-ax} \, dx$.

Do this integration by using integration by parts - twice.

Then $\displaystyle \sigma^2 = E(X^2) - \mu^2$.
• May 29th 2008, 10:18 PM
hatsoff

Quote:

Originally Posted by bloomsgal8
I'm working on a practice midterm and this question has me a bit stumped. Can anyone please help in working it out? Thank you!!

f(x) = ae^(-ax) with 0<equalto x < infinity

where a is constant, show that the mean and standard deviation are both 1/a

Thank you.

The mean value of x would be defined thusly:

$\displaystyle \int_{0}^{\bar{x}}ae^{-ax}dx=\int_{\bar{x}}^{\infty}ae^{-ax}dx$

Solving:

$\displaystyle [-e^{-ax}]_{0}^{\bar{x}}=[-e^{-ax}]_{\bar{x}}^{\infty}$

$\displaystyle -e^{-a\bar{x}}-(-e^{-a(0)})=[\lim_{x\to\infty}-e^{-ax}]-(-e^{-a\bar{x}})$

$\displaystyle -e^{-a\bar{x}}+1=0+e^{-a\bar{x}}$

$\displaystyle 1=2e^{-a\bar{x}}$

$\displaystyle \frac{1}{2}=e^{-a\bar{x}}$

$\displaystyle \ln{\frac{1}{2}}=\ln{e^{-a\bar{x}}}$

$\displaystyle -\ln{2}=-a\bar{x}\ln{e}$

$\displaystyle -\ln{2}=-a\bar{x}$

$\displaystyle \frac{\ln{2}}{a}=\bar{x}$

But you require the mean value of y. Can you figure it out from the above?
• May 29th 2008, 10:20 PM
mr fantastic
Quote:

Originally Posted by hatsoff
The mean value of x would be defined thusly:

$\displaystyle \int_{0}^{\bar{x}}ae^{-ax}dx=\int_{\bar{x}}^{\infty}ae^{-ax}dx$

Solving:

$\displaystyle [-e^{-ax}]_{0}^{\bar{x}}=[-e^{-ax}]_{\bar{x}}^{\infty}$

$\displaystyle -e^{-a\bar{x}}-(-e^{-a(0)})=[\lim_{x\to\infty}-e^{-ax}]-(-e^{-a\bar{x}})$

$\displaystyle -e^{-a\bar{x}}+1=0+e^{-a\bar{x}}$

$\displaystyle 1=2e^{-a\bar{x}}$

$\displaystyle \frac{1}{2}=e^{-a\bar{x}}$

$\displaystyle \ln{\frac{1}{2}}=\ln{e^{-a\bar{x}}}$

$\displaystyle -\ln{2}=-a\bar{x}\ln{e}$

$\displaystyle -\ln{2}=-a\bar{x}$

$\displaystyle \frac{\ln{2}}{a}=\bar{x}$

But you require the mean value of y. Mr F asks: Where has y come from?
[snip]

You've calculated the median value of X.
• May 29th 2008, 10:28 PM
hatsoff
Quote:

Originally Posted by mr fantastic
You've calculated the median value of X.

So I have, so I have.

Well, that's why I'm here--to get my fill of idiotic mistakes like that.

Quote:

Mr F asks: Where has y come from?
The x-y plane.