CDF of a function of 2 random variables

  • Thread starter vortmax
  • Start date
  • #1
vortmax
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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

[itex]f(T) = \lambda e^{-\lambda T}[/itex]
[itex] F(T) = 1 - e^{-\lambda T}[/itex]

Rayleigh-dist

[itex]f(T) = \frac{T}{\alpha^2} e^{\frac{-T}{2\alpha^2}}[/itex]
[itex]F(T) = 1 - e^{\frac{-T}{2\alpha^2}}[/itex]


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:

[itex] T = min(T_1,T_2)[/itex]

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

Answers and Replies

  • #2
sinat
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first time uses
 
  • #3
RPinPA
Science Advisor
Homework Helper
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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) ]##
 

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