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Math Help - Independent exponential random variables - find the joint probability distribution

  1. #1
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    Independent exponential random variables - find the joint probability distribution

    Let X and Y be independent exponential random variables with E(X)=2.5 and E(Y)=3.5.

    Find the joint probability distribution f(x,y).
    I know that the probability density function for X is \frac{2}{5}e^{\frac{-2x}{5} for x>0 and 0 otherwise.
    also
    I know that the probability density function for Y is \frac{2}{7}e^{\frac{-2x}{7} for x>0 and 0 otherwise.

    I am not sure about jointly. Do I simply multiply them together?

    Find P(X+Y>15).

    I have no idea how to begin this part, although I know the answer is .04.

    Can anyone help?
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  2. #2
    Super Member girdav's Avatar
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    Re: Independent exponential random variables - find the joint probability distributio

    Use the following result: if X and Y are two independent random variables which have a density, namely f for X and g for Y, then (X,Y) has a density F, which is given by F(x,y)=f(x)g(y).
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    Grand Panjandrum
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    Re: Independent exponential random variables - find the joint probability distributio

    Quote Originally Posted by CountingPenguins View Post
    Let X and Y be independent exponential random variables with E(X)=2.5 and E(Y)=3.5.

    Find the joint probability distribution f(x,y).
    I know that the probability density function for X is \frac{2}{5}e^{\frac{-2x}{5} for x>0 and 0 otherwise.
    also
    I know that the probability density function for Y is \frac{2}{7}e^{\frac{-2x}{7} for x>0 and 0 otherwise.

    I am not sure about jointly. Do I simply multiply them together?

    Find P(X+Y>15).

    I have no idea how to begin this part, although I know the answer is .04.

    Can anyone help?
    The definition of independence is that

    f(x,y)=g(x)h(y)

    CB
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Independent exponential random variables - find the joint probability distributio

    Quote Originally Posted by CountingPenguins View Post
    Let X and Y be independent exponential random variables with E(X)=2.5 and E(Y)=3.5...

    ... find P(X+Y>15)...

    I have no idea how to begin this part, although I know the answer is .04...
    About the second question we have two independent exponential random variables X and Y ande their p.d.f are...

    f_{x}(t)= \frac{2}{5}\ e^{-\frac {2}{5} t}\ \mathcal {U} (t) (1)

    f_{y}(t)= \frac{2}{7}\ e^{-\frac {2}{7} t}\ \mathcal{U} (t) (2)

    ...where \mathcal {U} (t) is the so called 'Heaviside step function' ...

    \mathcal {U} (t)= \begin{cases}1&t>0\\\frac{1}{2}&t=0\\0&t<0\end{cas  es}

    Now if we set Z=X+Y the p.d.f. of Z is given by the convolution of f_{x}(t) and f_{y}(t) ...

    f_{z}(t)= f_{x}(t)*f_{y}(t) = \int_{0}^{t} f_{x}(\tau)\ f_{y}(t-\tau)\ d \tau (4)

    Using a basic property of the Laplace Transform we have that...

    \mathcal{L} \{f_{z}(t)\}= \mathcal{L} \{f_{x}(t)\}\ \mathcal{L} \{f_{y}(t)\} (5)

    ... so that...

    \mathcal{L} \{f_{x}(t)\}= \frac{\frac{2}{5}}{s + \frac{2}{5}} (6)

    \mathcal{L} \{f_{y}(t)\}= \frac{\frac{2}{7}}{s + \frac{2}{7}} (7)

    \mathcal{L} \{f_{z}(t)\}= \frac{\frac{4}{35}}{(s + \frac{2}{5})\ (s + \frac{2}{7})} = \frac{1}{s + \frac{2}{7}} -  \frac{1}{s + \frac{2}{5}} (8)

    ... and from (8) taking the inverse Laplace Tranform...

    f_{z} (t) = e^{-\frac{2}{7} t} - e^{-\frac{2}{5} t} (9)

    Once You have the p.d.f. of Z You can obtain the requested probability by integration...

    P\{Z>15\}= \int_{15}^{\infty} (e^{-\frac{2}{7} t} - e^{-\frac{2}{5} t})\ dt = \frac{7}{2}\ e^{-\frac{30}{7}} - \frac{5}{2}\ e^{-6} =.041976373124...  (9)

    Kind regards

    \chi \sigma
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