Probability about three independent exponential random variables

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.

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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
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

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.

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this is quite clever. –  geraldgreen Nov 16 '11 at 17:02

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 $\frac{1}{15}$.

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nice use of memorylessness. –  geraldgreen Nov 11 '11 at 20:47

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}. }

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The naive way would be to use conditional probability:

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

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