# Independent exponential random variables - find the joint probability distribution

• Aug 12th 2011, 08:08 AM
CountingPenguins
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?
• Aug 13th 2011, 07:09 AM
girdav
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 $\displaystyle F(x,y)=f(x)g(y)$.
• Aug 13th 2011, 09:28 AM
CaptainBlack
Re: Independent exponential random variables - find the joint probability distributio
Quote:

Originally Posted by CountingPenguins
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
• Aug 13th 2011, 12:10 PM
chisigma
Re: Independent exponential random variables - find the joint probability distributio
Quote:

Originally Posted by CountingPenguins
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

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