Multi-variable pdf.

**Mathlv** · December 2nd 2011, 08:27 PM

Hello,
Would someone please help explain this to me?
Given:
$f(x,y) = {\rm{ \{ }}{e^{ - y}},{\rm{ }}0 < x < y$ and = 0 elsewhere
Find P(x>1, y>1)
So the set up happen to be as follow and then solve the integration
$\int_1^\infty {\int_1^y {{e^{ - y}}} dxdy}$

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

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

**mr fantastic** · December 2nd 2011, 09:35 PM

Originally Posted by Mathlv

Hello,
Would someone please help explain this to me?
Given:
$f(x,y) = {\rm{ \{ }}{e^{ - y}},{\rm{ }}0 < x < y$ and = 0 elsewhere
Find P(x>1, y>1)
So the set up happen to be as follow and then solve the integration
$\int_1^\infty {\int_1^y {{e^{ - y}}} dxdy}$

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

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

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

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

Your instructor expects you to know how to integrate.

**alexmahone** · December 2nd 2011, 10:11 PM

Originally Posted by mr fantastic

Integral equals 1 because $\lim_{\alpha \to +\infty} [-e^{-u}]_0^{+\infty} = 1$

$\lim_{\alpha \to +\infty} [-e^{-u}]_0^{\alpha} = 1$

**mr fantastic** · December 3rd 2011, 12:31 PM

Originally Posted by alexmahone

$\lim_{\alpha \to +\infty} [-e^{-u}]_0^{\alpha} = 1$

Indeed. A minor typo. My mistake.

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