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Thread: 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 $\displaystyle E(X)=2.5$ and $\displaystyle E(Y)=3.5.$

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

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

    Find $\displaystyle P(X+Y>15)$.

    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 $\displaystyle X$ and $\displaystyle Y$ are two independent random variables which have a density, namely $\displaystyle f$ for $\displaystyle X$ and $\displaystyle g$ for $\displaystyle Y$, then $\displaystyle (X,Y)$ has a density $\displaystyle F$, which is given by $\displaystyle F(x,y)=f(x)g(y)$.
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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 $\displaystyle E(X)=2.5$ and $\displaystyle E(Y)=3.5.$

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

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

    Find $\displaystyle P(X+Y>15)$.

    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

    $\displaystyle f(x,y)=g(x)h(y)$

    CB
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    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 $\displaystyle E(X)=2.5$ and $\displaystyle E(Y)=3.5$...

    ... find $\displaystyle P(X+Y>15)$...

    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...

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

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

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

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

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

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

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

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

    ... so that...

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

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

    $\displaystyle \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...

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

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

    $\displaystyle 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

    $\displaystyle \chi$ $\displaystyle \sigma$
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