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Math Help - Multi-variable pdf.

  1. #1
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    Multi-variable pdf.

    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  {}
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  2. #2
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    Re: Probability Problem

    Quote Originally Posted by Mathlv View Post
    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.
    Last edited by mr fantastic; December 3rd 2011 at 12:30 PM. Reason: Corrected minor typo. Thankyou alexmahone.
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Probability Problem

    Quote Originally Posted by mr fantastic View Post
    Integral equals 1 because \lim_{\alpha \to +\infty} [-e^{-u}]_0^{+\infty} = 1
    \lim_{\alpha \to +\infty} [-e^{-u}]_0^{\alpha} = 1
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  4. #4
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    Re: Probability Problem

    Quote Originally Posted by alexmahone View Post
    \lim_{\alpha \to +\infty} [-e^{-u}]_0^{\alpha} = 1
    Indeed. A minor typo. My mistake.
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