# Thread: Expected value of X

1. ## Expected value of X

I am trying to find the expected value of X.

$
f(x) = 2x e^{ - x^2 } dx,
for all x \in R = (0,\infty )
$

$
u = E[X] = \int_R^{} {x*f(x)} dx = 2\int\limits_0^\infty {x^2 e^{ - x^2 } dx}
$

I could not find any integral table to make this any easier and my trusty ti-92+ would not integrate this either.

So, I have tried to solve by letting
$
u = x^2 and\_du = 2xdx
$

Again, I have gotten nowhere.

Any suggestions would be appreciated.

Thanks.

Note:
I just found a table that lists the integral = .5*(pi)^1/2.

2. Originally Posted by kid funky fried
I am trying to find the expected value of X.

$
f(x) = 2x e^{ - x^2 } dx,
for all x \in R = (0,\infty )
$

$
u = E[X] = \int_R^{} {x*f(x)} dx = 2\int\limits_0^\infty {x^2 e^{ - x^2 } dx}
$

I could not find any integral table to make this any easier and my trusty ti-92+ would not integrate this either.

So, I have tried to solve by letting
$
u = x^2 and\_du = 2xdx
$

Again, I have gotten nowhere.

Any suggestions would be appreciated.

Thanks.

Note:
I just found a table that lists the integral = .5*(pi)^1/2.
It can be reduced to a 'standard' definite integral using integration by parts. I have no time now but will post later.

3. Originally Posted by mr fantastic
It can be reduced to a 'standard' definite integral using integration by parts. I have no time now but will post later.
You might find this thread has what you want: http://www.mathhelpforum.com/math-he...-function.html

After reading this, let me know if you still need help with your specific calculation.

4. Mr. Fantastic,

So, to answer your question, i guess i am lost...i have given this problem hours of thought and feel like i have gotten nowhere.
this is a homework problem and many classmates and i are baffled.
i will review the thread again to see if i can understand how it applies.

5. Originally Posted by kid funky fried
Mr. Fantastic,

So, to answer your question, i guess i am lost...i have given this problem hours of thought and feel like i have gotten nowhere.
this is a homework problem and many classmates and i are baffled.
i will review the thread again to see if i can understand how it applies.
$E(X) = 2 \int_{0}^{+\infty} x^2 e^{-x^2} \, dx$.

Substitute $u = x^2 \Rightarrow dx = \frac{du}{dx}$:

$E(X) = \int_{0}^{+\infty} u^{1/2} e^{-u} \, du$.

There is a well known special function called the Gamma function:

Definition: $\Gamma(n) = \int_{0}^{+\infty} u^{n-1} e^{-u} \, du$.

Special property: $\Gamma (n+1) = n \Gamma (n)$.

Therefore $E(X) = \Gamma \left(\frac{3}{2}\right) = \frac{1}{2} \Gamma \left(\frac{1}{2}\right)$.

It's well known that $\Gamma \left(\frac{1}{2}\right) = \sqrt{\pi}$.

Therefore $E(X) = \frac{1}{2} \, \sqrt{\pi}$.