# CDF of a function of 2 random variables

• vortmax
The events (T1 <= x and T2 <= T1) and (T2 <= x and T1 <= T2) are independent (why?), so the P(... OR ...) above is equal to the sum of the P(...)'s.The first P(...) is a bit easier to calculate: it's just ##P(T1 <= x) * P(T2 <= T1)##. The second P(...) you can calculate by symmetry: it's equal to the first one, because T1 and T2 are completely interchangeable.So you get##P(T \leq x) = P(T1 <= x) * P(T2 <= T1) + P(T2 <= x) * P(T1 <=

## Homework Statement

Two toys are started at the same time each with a different battery. The first battery has a lifetime that is exponentially distributed with mean 100 min; the second battery has a lifetime that is Rayleigh-distributed with a mean 100 minutes.

a) Find the CDF to the time T until the battery in a toy first runs out
b) Suppose that both toys are still operational at 100 minutes. Find the CDF of the time T2 that subsequently elapses until the battery in a toy first runs out
c) in part b, find the cdf to the total time that elapses until a battery first fails.

## Homework Equations

Exponential Dist

$f(T) = \lambda e^{-\lambda T}$
$F(T) = 1 - e^{-\lambda T}$

Rayleigh-dist

$f(T) = \frac{T}{\alpha^2} e^{\frac{-T}{2\alpha^2}}$
$F(T) = 1 - e^{\frac{-T}{2\alpha^2}}$

## The Attempt at a Solution

First, I calculated values for lambda and alpha based on the means ... but this part isn't entirely necessary to the final solution.

a) I need to find the cdf of the function:

$T = min(T_1,T_2)$

where $T_1$ and $T_2$ are the two RV's respectively... but I'm really at a loss about how to proceed.

first time uses

This is a special case of what is called "order statistics", the lowest, 2nd lowest, 3rd lowest, of a set of random variables. You can google on that keyword to learn more about how they're calculated.

##T \leq x## means either T1 is the smallest and is ##\leq x##, or T2 is the smallest and is ##\leq x##.

So ##P(T \leq x) = P[ (T1 \leq x \text{ and } T2 <= T1) \text{ OR } (T2 \leq x \text{ and } T1 \leq T2) ]##