# Multi-variable pdf.

• Dec 2nd 2011, 08:27 PM
Mathlv
Multi-variable pdf.
Hello,
Given:
$\displaystyle f(x,y) = {\rm{ \{ }}{e^{ - y}},{\rm{ }}0 < x < y$ and = 0 elsewhere
Find P(x>1, y>1)
So the set up happen to be as follow and then solve the integration
$\displaystyle \int_1^\infty {\int_1^y {{e^{ - y}}} dxdy}$

= $\displaystyle \int_1^\infty {{e^{ - y}}(y - 1)dy}$

Let u = y-1 $\displaystyle \Rightarrow$ du = dy
= $\displaystyle \int_0^\infty {{e^{ - (u + 1)}}udu = {e^{ - 1}}} \int_0^\infty {u{e^{ - u}}} du$
= $\displaystyle {e^{ - 1}} \times 1 = {e^{ - 1}}$
What I do not understand is how my instructor comes to $\displaystyle {e^{ - 1}} \times 1 = {e^{ - 1}}$
How did he gets to 1?
Also I do not know how he changes from $\displaystyle \int_1^\infty {}$ to $\displaystyle \int_0^\infty {}$
• Dec 2nd 2011, 09:35 PM
mr fantastic
Re: Probability Problem
Quote:

Originally Posted by Mathlv
Hello,
Given:
$\displaystyle f(x,y) = {\rm{ \{ }}{e^{ - y}},{\rm{ }}0 < x < y$ and = 0 elsewhere
Find P(x>1, y>1)
So the set up happen to be as follow and then solve the integration
$\displaystyle \int_1^\infty {\int_1^y {{e^{ - y}}} dxdy}$

= $\displaystyle \int_1^\infty {{e^{ - y}}(y - 1)dy}$

Let u = y-1 $\displaystyle \Rightarrow$ du = dy
= $\displaystyle \int_0^\infty {{e^{ - (u + 1)}}udu = {e^{ - 1}}} \int_0^\infty {u{e^{ - u}}} du$
= $\displaystyle {e^{ - 1}} \times 1 = {e^{ - 1}}$
What I do not understand is how my instructor comes to $\displaystyle {e^{ - 1}} \times 1 = {e^{ - 1}}$
How did he gets to 1?
Also I do not know how he changes from $\displaystyle \int_1^\infty {}$ to $\displaystyle \int_0^\infty {}$

Integral terminals: Change of variable u = y - 1. y = 1 => u = 0, y = +oo => u = +oo.

Integral equals 1 because $\displaystyle \lim_{\alpha \to +\infty} [-e^{-u}]_0^{\alpha} = 1$. NB: The integral is improper hence the use of the limit.

Your instructor expects you to know how to integrate.
• Dec 2nd 2011, 10:11 PM
alexmahone
Re: Probability Problem
Quote:

Originally Posted by mr fantastic
Integral equals 1 because $\displaystyle \lim_{\alpha \to +\infty} [-e^{-u}]_0^{+\infty} = 1$

$\displaystyle \lim_{\alpha \to +\infty} [-e^{-u}]_0^{\alpha} = 1$
• Dec 3rd 2011, 12:31 PM
mr fantastic
Re: Probability Problem
Quote:

Originally Posted by alexmahone
$\displaystyle \lim_{\alpha \to +\infty} [-e^{-u}]_0^{\alpha} = 1$

Indeed. A minor typo. My mistake.