# Conditional expectation of three exponential distributed r.v.

## Main Question or Discussion Point

I've been struggling with this problem for more than 4 days now:

Let A, B and C be exponential distributed random variables with parameters lambda_A, lambda_B and lambda_C, respectively.

Calculate E [ B | A < B < C ] in terms of the lambda's.

I always seem get an integral which is impossible to calculate...

Who has a suggestion how to solve this problem?

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I always seem get an integral which is impossible to calculate...
That's how I'd try to do it too, can you show us what you got and where you were stuck?

First a remark: we may assume that A, B and C are independent random variables (this may help a lot).

$$E [ B | A < B < C ] = \int_{0}^\infty E [ B \, | \, A < B < C, A = a ] \; f_{A} ( a ) \; da$$
$$= \int_{0}^\infty \int_{0}^\infty E [ B \, | \, A < B < C, A = a, C=c ] \; f_{A} ( a ) \; f_{C} ( c ) \; da \; dc.$$
(Along with all that follows, this integral is only to be considered over R_+^2 and for c>a.)
$$= \int_{0}^\infty \int_{0}^\infty E [ B \, | \, a < B < c, A = a, C=c ] \; f_{A} ( a ) \; f_{C} ( c ) \; da \; dc$$
(by independence)
$$= \int_{0}^\infty \int_{0}^\infty E [ B \, | \, a < B < c ] \; f_{A} ( a ) \; f_{C} ( c ) \; da \; dc$$

We want to compute $$E [ B \, | \, a < B < c ].$$

First note that
$$E [ B \, | \, a < B < c ] = \frac{ \int_a^c t f_{B} (t) dt }{P( B \in [a,c])}.$$

We have$$\int_a^c t f_{B} (t) dt = ... = (a e^{- \lambda_B a} - c e^{- \lambda_B c}) + \frac{1}{\lambda_B} ( e^{- \lambda_B a} - e^{- \lambda_B c} )$$
and
$$P(B \in [a,c]) = F_B (c) - F_B (a) = ... = e^{- \lambda_B a} - e^{- \lambda_B c}.$$

Hence
$$E [ B \, | \, a < B < c ] = \frac{ (a e^{- \lambda_B a} - c e^{- \lambda_B c}) + \frac{1}{\lambda_B} ( e^{- \lambda_B a} - e^{- \lambda_B c} ) }{e^{- \lambda_B a} - e^{- \lambda_B c}},$$
$$E [ B \, | \, a < B < c ] = \frac{ a e^{- \lambda_B a} - c e^{- \lambda_B c} }{e^{- \lambda_B a} - e^{- \lambda_B c}} + \frac{1}{\lambda_B}.$$

So
$$E [ B | A < B < C ] = \int_{0}^\infty \int_{0}^\infty \left( \frac{ a e^{- \lambda_B a} - c e^{- \lambda_B c} }{e^{- \lambda_B a} - e^{- \lambda_B c}} + \frac{1}{\lambda_B} \right) \; f_{A} ( a ) \; f_{C} ( c ) \; da \; dc$$
$$= \int_{0}^\infty \int_{0}^\infty \left( \frac{ a e^{- \lambda_B a} - c e^{- \lambda_B c} }{e^{- \lambda_B a} - e^{- \lambda_B c}} \right) \lambda_A e^{- \lambda_A a} \; \lambda_C e^{- \lambda_C c} \; da \; dc +$$
$$\int_{0}^\infty \int_{0}^\infty \frac{1}{\lambda_B} \lambda_A e^{- \lambda_A a} \; \lambda_C e^{- \lambda_C c} \; da \; dc$$

Here's my problem: I can't compute the first of these integrals... And even then I'm not sure if what I'm doing here is correct... Is my calculation right, or is it wrong!? Is there perhaps a better way to calculate this? Etc.

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$$E [ B | A < B < C ] = \int_{0}^\infty E [ B \, | \, A < B < C, A = a ] \; f_{A} ( a ) \; da$$
I'm not sure about this line - the marginals of the conditional distribution aren't necessarily exponential. Another approach is to apply directly the definition of conditional probabilities, so that

$$E[B|A<B<C] = E[B.I(A<B<C)]/P[A<B<C]$$

which is a ratio of integrals that should both be tractable.

Hope this helps.

I'm not sure about this line - the marginals of the conditional distribution aren't necessarily exponential. Another approach is to apply directly the definition of conditional probabilities, so that

$$E[B|A<B<C] = E[B.I(A<B<C)]/P[A<B<C]$$

which is a ratio of integrals that should both be tractable.

Hope this helps.
Isn't this just the "law of total expectation"?

I don't really understand your point I'm afraid.

EDIT: Yes, we can also go by the second approach.

Then how to calculate $$E[B.I(A<B<C)]$$ exactly ? (I'm not sure how to start with this integral.)

Isn't this just the "law of total expectation"?

I don't really understand your point I'm afraid.

EDIT: Yes, we can also go by the second approach.

Then how to calculate $$E[B.I(A<B<C)]$$ exactly ? (I'm not sure how to start with this integral.)
I now understand what you mean by "I'm not sure about this line (...)": you're completely right, it's wrong!

Then how to calculate $$E[B.I(A<B<C)]$$ exactly ? (I'm not sure how to start with this integral.)
Ok, this can be simplified to

$$\int_R b.f_A(a).f_B(b).f_C(c)da.db.dc$$

where R is the region a<b<c.