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Thread: A continous random variable...

  1. #1
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    Joint probability density function

    Question
    The joint probability density function of X and Y is given by

    f(x,y) = $\displaystyle \begin{cases} exp[-(x+y)] & 0 \le x < \infty , 0 \le y < \infty \\
    0 & otherwise \end{cases}$

    Compute E[X] and P[Y>2]
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  2. #2
    MHF Contributor matheagle's Avatar
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    They are indep by inspection.
    Both X and Y are exponential with mean 1.
    So E(X)=1 and

    $\displaystyle P(Y>2)=\int_2^{\infty}e^{-y}dy=e^{-2}$
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  3. #3
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    I am stuck

    Quote Originally Posted by matheagle View Post
    They are indep by inspection.
    Both X and Y are exponential with mean 1.
    So E(X)=1 and

    $\displaystyle P(Y>2)=\int_2^{\infty}e^{-y}dy=e^{-2}$

    Here is what i have done


    g(x) = $\displaystyle \int_{0}^{ \infty}$ f(x,y) dy
    = $\displaystyle \int_{0}^{ \infty} $ exp[-(x+y)] = -$\displaystyle \int_{0}^{ \infty}$ exp(x+y) dy
    = - $\displaystyle \int_{0}^{ \infty}$ $\displaystyle exp(x) \ . \ exp(y) dy$

    = -exp(x) $\displaystyle \int_{0}^{ \infty}$ exp(y) dy = ?_________I am stuck here what is exp(y)....
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  4. #4
    MHF Contributor matheagle's Avatar
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    INFINITY since you forgot the negative in the exponent.
    DID you read my previous post?
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  5. #5
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    I m stuck

    Quote Originally Posted by matheagle View Post
    INFINITY since you forgot the negative in the exponent.
    DID you read my previous post?

    I have read ur prev post, but what about the -ve sign ..........i have not forgotten about it but i feel i am missing something ............
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  6. #6
    MHF Contributor matheagle's Avatar
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    I'll fix this, but there is to reason to do it this way...


    $\displaystyle g(x) = e^{-x}\int_{0}^{ \infty}e^{-y}dy=e^{-x} $

    you should know that the derivative and integral of $\displaystyle e^{-y}$ is $\displaystyle -e^{-y}$
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  7. #7
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    Now i am again stuck

    Quote Originally Posted by matheagle View Post
    I'll fix this, but there is to reason to do it this way...


    $\displaystyle g(x) = e^{-x}\int_{0}^{ \infty}e^{-y}dy=e^{-x} $

    you should know that the derivative and integral of $\displaystyle e^{-y}$ is $\displaystyle -e^{-y}$
    g(x) = $\displaystyle e^{-x}$ or $\displaystyle -e^x$.......which one is it?????
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  8. #8
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    Quote Originally Posted by zorro View Post
    g(x) = $\displaystyle e^{-x}$ or $\displaystyle -e^x$.......which one is it?????
    Why are you confused? Matheagle's post says clearly that $\displaystyle g(x) = e^{-x}$. You should be able to calculate $\displaystyle \int_{0}^{ \infty}e^{-y} \, dy = 1$ .... (Besides, a pdf cannot be negative and $\displaystyle -e^{x} < 0$ for all real values of x ....!!)
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  9. #9
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    Thank u so much

    Quote Originally Posted by mr fantastic View Post
    Why are you confused? Matheagle's post says clearly that $\displaystyle g(x) = e^{-x}$. You should be able to calculate $\displaystyle \int_{0}^{ \infty}e^{-y} \, dy = 1$ .... (Besides, a pdf cannot be negative and $\displaystyle -e^{x} < 0$ for all real values of x ....!!)
    Thanks u so much Mr fantastic and matheagle.........Zorro will never forget it .........Its because of u i am able to wave my sword in the air ....On gourd.......
    cheers mate
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