Probability with two random variables

**Justin Lo** · July 24th 2009, 05:11 AM

Consider two independent random variables: $r,s$ , where
$r\sim f(r)$ and $s \sim U[0,x]$ .
( $U[0,x]$ means uniform within support $0$ and $x$ .)

What is the following probability:

$\Pr (r-ts>p)$ , where $t$ and $p$ are constants.

Any suggestions, tips or ideas?

**Moo** · July 24th 2009, 06:42 AM

Hello,

Originally Posted by Justin Lo

Consider two independent random variables: $r,s$ , where
$r\sim f(r)$ and $s \sim U[0,x]$ .
( $U[0,x]$ means uniform within support $0$ and $x$ .)

What is the following probability:

$\Pr (r-ts>p)$ , where $t$ and $p$ are constants.

Any suggestions, tips or ideas?

Since they're independent, their joint probability density function is $f(r)\cdot \frac 1x$

Now, $\mathbb{P}(r-ts>p)=\mathbb{E}\left(\mathbf{1}_{r-ts>p}\right)$ (1 is the indicator function)

By the law of the unconscious statistician, which states that for any measurable function f, $\mathbb{E}(f(X_1,\dots,X_n))=\int_{\mathbb{R}^n} f(x_1,\dots,x_n) g(x_1,\dots,x_n) ~dx_1\dots dx_n$

where g is the joint probability function of $(X_1,\dots,X_n)$

So here, we have :

$\mathbb{P}(r-ts>p)=\int_0^x\int_{\mathbb{R}} \frac{f(r)}{x} \cdot \mathbf{1}_{r-ts>p} ~dr ~ds$

now, $\mathbf{1}_{r-ts>p}$ has to be interpreted as a region.
Since we first integrate with respect to r, consider it with respect to r :
$r>p-ts$

So finallly :

$\mathbb{P}(r-ts>p)=\int_0^x\int_{p-ts}^\infty \frac{f(r)}{x} ~dr ~ds$

Note : the inner integral doesn't necessarily goes to infinity, it depends on the support of the rv r, which will be expressed in the function f.

Thread: Probability with two random variables

LinkBack

Thread Tools

Display

Probability with two random variables

Similar Math Help Forum Discussions

Random variables and probability distributions #1

Probability of Random Variables

Probability - random variables

Join Probability & Random Variables

probability with random variables

Search Tags