Independent exponential random variables

**CountingPenguins** · August 12th 2011, 08:08 AM

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?

**girdav** · August 13th 2011, 07:09 AM

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 $F(x,y)=f(x)g(y)$ .

**CaptainBlack** · August 13th 2011, 09:28 AM

Originally Posted by CountingPenguins

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

**chisigma** · August 13th 2011, 12:10 PM

Originally Posted by CountingPenguins

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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