# Clearification on Exponential Distribution

• Nov 23rd 2008, 02:37 PM
lllll
Clearification on Exponential Distribution
Assume $X_1$ and $X_2$ are independent random variables, both following a exponential distribution. Calculate $P\{X_1

The way it's done in the book

$P\{X_1 < X_2\} = \int_0^\infty P\{X_1 < X_2|X_1 = x\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ (a)

$= \int_0^\infty P\{ x< X_2\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ (b)

$= \int_0^\infty e^{-\lambda_2x}{\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ (c)

I'm wondering is why are they putting the pdf of exponential function (the part in blue)?

in part (b) would is it $\int_0^\infty \frac{P\{X_1 < X_2, X=x\}}{P\{X_1=x\}} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ $\Longrightarrow \int_0^\infty \frac{P\{X_1 < X_2\} P\{X=x\}}{P\{X_1=x\}} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ $\Longrightarrow \int_0^\infty P\{ x< X_2\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$

in part (c) isn't it $P\{ x< X_2\} = \lambda_2 e^{-\lambda_2x}$, therefore wouldn't it be: $\int_0^\infty {\color [cmyk] {0.75,0,0.75,0.45}\lambda_2}e^{-\lambda_2x}{\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$

why isn't there a ${\color [cmyk] {0.75,0,0.75,0.45}\lambda_2}$?
• Nov 23rd 2008, 06:41 PM
mr fantastic
Quote:

Originally Posted by lllll
Assume $X_1$ and $X_2$ are independent random variables, both following a exponential distribution. Calculate $P\{X_1

The way it's done in the book

$P\{X_1 < X_2\} = \int_0^\infty P\{X_1 < X_2|X_1 = x\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ (a)

$= \int_0^\infty P\{ x< X_2\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ (b)

$= \int_0^\infty e^{-\lambda_2x}{\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ (c)

I'm wondering is why are they putting the pdf of exponential function (the part in blue)?

in part (b) would is it $\int_0^\infty \frac{P\{X_1 < X_2, X=x\}}{P\{X_1=x\}} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ $\Longrightarrow \int_0^\infty \frac{P\{X_1 < X_2\} P\{X=x\}}{P\{X_1=x\}} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ $\Longrightarrow \int_0^\infty P\{ x< X_2\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$

in part (c) isn't it $P\{ x< X_2\} = \lambda_2 e^{-\lambda_2x}$, therefore wouldn't it be: $\int_0^\infty {\color [cmyk] {0.75,0,0.75,0.45}\lambda_2}e^{-\lambda_2x}{\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$

why isn't there a ${\color [cmyk] {0.75,0,0.75,0.45}\lambda_2}$?

Here is how I would do it:

The joint pdf of $X_1$ and $X_2$ is $f(x_1, x_2) = \lambda_1 \lambda_2 e^{-\lambda_1 x_1} e^{-\lambda_2 x_2}$.

Then $\Pr(X_1 < X_2) = \int_{x_2 = 0}^{x_2 = +\infty} \int_{x_1 = 0}^{x_1 = x_2} f(x_1, x_2) \, dx_1 \, dx_2 = \, ....$
• Nov 23rd 2008, 10:38 PM
lllll
just out of curiosity, if you were to add a third variable $X_3$ which is also independent and follows an exponential distribution, and wanted to calculate

$P\{X_1 would you simply have a triple integral such as:

$P\{X_1

where $f(x_1,x_2,x_3) = \lambda_1 e^{-\lambda_1 x_1} \cdot \lambda_2 e^{-\lambda_2 x_2} \cdot \lambda_3 e^{-\lambda_3 x_3}$ ?
• Nov 23rd 2008, 10:41 PM
mr fantastic
Quote:

Originally Posted by lllll
just out of curiosity, if you were to add a third variable $X_3$ which is also independent and follows an exponential distribution, and wanted to calculate

$P\{X_1 would you simply have a triple integral such as:

$P\{X_1

where $f(x_1,x_2,x_3) = \lambda_1 e^{-\lambda_1 x_1} \cdot \lambda_2 e^{-\lambda_2 x_2} \cdot \lambda_3 e^{-\lambda_3 x_3}$ ?

I'd calculate it as $\Pr(X_2 < X_3) - \Pr(X_2 < X_1)$.
• Dec 12th 2008, 06:58 AM
Laurent
Quote:

Originally Posted by mr fantastic
I'd calculate it as $\Pr(X_2 < X_3) - \Pr(X_2 < X_1)$.

Not quite, since this even can be negative... It should be:
$P(X_1 $=P(X_2
and it is well known that $\min(X_1,X_3)$ is an exponential random variable of parameter $\lambda_1+\lambda_3$, and it is independent of $X_2$ so, in application of the result of your initial question,
$P(X_1

One easily checks that the integral gives the same result. By the way, the computation of the integral gives by induction the general formula:
$P(X_1<\cdots
• Dec 12th 2008, 12:38 PM
mr fantastic
Quote:

Originally Posted by Laurent
Not quite, since this even can be negative... It should be:
$P(X_1 $=P(X_2
and it is well known that $\min(X_1,X_3)$ is an exponential random variable of parameter $\lambda_1+\lambda_3$, and it is independent of $X_2$ so, in application of the result of your initial question,
$P(X_1

One easily checks that the integral gives the same result. By the way, the computation of the integral gives by induction the general formula:
$P(X_1<\cdots

Ha. Thanks for the catch. Obvious in hindsight, I'm annoyed I missed it the first time.