Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose we have three independent exponential random variables $A$, $B$ and $C$ with respective parameters $1$, $2$ and $3$.

Calculate $P(A<B<C)$.

The hint says this problem could be solved with calculus and without calculus. I am really curious how to approach it with different methods.

share|improve this question
1  
As an aside, here's another probability question answered with the same formula: math.stackexchange.com/questions/4220/… –  Byron Schmuland Nov 12 '11 at 0:19
2  
There are two conventional parametrizations of the exponential distribution. Which one do you have in mind? Do you mean $B$ has an exponential distribution with rate $2$ and therefore expectation $1/2$, or do you mean $B$ has an exponential distribution with expectation $2$? In the former case the density is $f(x) = 2e^{-2x}$ for $x>0$; in the latter it is $(1/2)e^{-x/2}$ for $x>0$. –  Michael Hardy Nov 12 '11 at 3:40
add comment

4 Answers

up vote 5 down vote accepted

It is rather well known that $$P(A<B)=\frac{\lambda_a}{\lambda_a+\lambda_b}.$$

You should also know the fact that $$\min(A,B)\sim \mathcal E(\lambda_a+\lambda_b),$$ think about the first arrival time of the addition of two Poisson processes, for example.

From there, you could decompose the event $\{A<B<C\}$ into the intersection of the two independent events $$\{A<\min(B,C)\}$$ and $$\{B<C\}$$ to get an intuition of the result.

share|improve this answer
    
this is quite clever. –  geraldgreen Nov 16 '11 at 17:02
add comment

Here is how to solve it without much calculus. Remember that for exponential variable $X$ with parameter $\lambda$, $\mathbb{P}(X > x) = \exp(-\lambda x)$ for $x \ge 0$.

We will make use of the following result, that relies on memoryless property of exponential distribution, i.e. $\mathbb{E}(f(X-a) ; X > a) = \mathbb{E}(f(X))$: $$ \begin{eqnarray} \mathbb{E}( \mathbf{1}_{X > a} \exp(-\mu X)) &=& \mathbb{E}(\exp(-\mu X); X > a) \mathbb{P}(X > a) \\ &=& \exp(-\mu a) \cdot \mathbb{E}(\exp(-\mu (X-a)); X > a) \cdot \exp(-\lambda a) \\ &=& \exp(-(\mu+\lambda) a) \cdot \mathbb{E}(\exp(-\mu X) ) \\ &=& \exp(-(\mu+\lambda) a) \cdot \frac{\lambda}{\lambda+\mu} \end{eqnarray} $$

Now, use conditioning: $$ \begin{eqnarray} \mathbb{P}(C > B > A) &=& \mathbb{E}_A( \mathbb{E}_B( \mathbf{1}_{B>a} \mathbb{P}(C > b ; B=b, A=a) ; A=a)) \\ &=& \mathbb{E}_A( \mathbb{E}_B( \mathbf{1}_{B>a} \exp(-\lambda_c b) ; A=a)) \\ &=& \mathbb{E}_A( \frac{\lambda_b}{\lambda_b+\lambda_c} \exp(-a(\lambda_b+\lambda_c) ) \\ &=& \frac{\lambda_b}{\lambda_b+\lambda_c} \cdot \frac{\lambda_a}{\lambda_a + \lambda_b+\lambda_c} \end{eqnarray} $$

With $\lambda_a=1$, $\lambda_b = 2$, $\lambda_c=3$, the answer comes to $\frac{1}{15}$.

share|improve this answer
    
nice use of memorylessness. –  geraldgreen Nov 11 '11 at 20:47
add comment

This can be done from first principles using minimal Calculus (just differentiation and integration of monomials). By one definition, when $X$ has an Exponential distribution with parameter $\lambda\gt 0$, $\Pr(X\le x) = 1 - \exp(-\lambda x)$. Therefore, for any $0\lt x\le 1$,

$$\Pr(\exp(-X)\lt x) = \Pr(X \gt -\log(x)) = \exp(\lambda\log(x)) = x^\lambda.$$

By differentiation we find the probability density of $\exp(-X)$ is $\lambda x^{\lambda-1}\ dx$. The solution is now obtained by integrating the joint density of $(A,B,C)$ over the event $A\lt B\lt C$ via this transformation:

$$\eqalign{ \Pr(A \lt B \lt C) &= \Pr(\exp(-C)\lt \exp(-B)\lt \exp(-A)) \\ &= \int_0^1 da \int_0^a 2b\ db\int_0^b 3c^2\ dc \\ &= \int_0^1 da \int_0^a 2b\ b^3\ db\\ &= \frac{2}{5}\int_0^1 a^5\ da \\ &= \frac{1}{15}. }$$

share|improve this answer
add comment

The naive way would be to use conditional probability:

$\mathrm P(A<B<C) = \int_0^\infty \mathrm P(A < B < x) f_C(x) \mathrm d x = \int_0^\infty \int_0^x \mathrm P(A<y) f_B(y) f_C(x) \mathrm d y \mathrm d x$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.