definite integral over random interval

[benjaminmar8]

Hi, all,

assuming a and b are random variables and their pdf f(a) and f(b) are known. then, how do I solve for the definite integral given as v=\int\limits_{a}^{b} g(x) dx, where g(x) is a function of x? or, how do I solve the pdf of v?

Thanks a lot..

[EnumaElish]

I think the answer is "use the fundamental theorem of calculus." Operationally, suppose each of a and b is Bernoulli with probabilities p and q: Pr[a = a1] = 1 - Pr[a = a2] = p, Pr[b = b1] = 1 - Pr[b = b2] = q.

Then:
v = G[b1] - G[a1] with prob = pq,
v = G[b2] - G[a1] with prob. p(1-q)
v = G[b1] - G[a2] with prob. (1-p)q
v = G[b2] - G[a2] with prob (1-p)(1-q)

where "the integral of g from a to b" = G[b] - G[a].

[bpet]

Another approach is to first work out the cdf P[v<=y]. To do this (assuming a and b are independent) it might be helpful to sketch the 2d region of points (a,b) where G[b]-G[a]<=y.

[winterfors]

If we define that
y(a,b)=\int\limits_{b=-\infty}^{\infty} \int\limits_{a}^{b} g(x) dx
we can state the pdf over the joint space of variables a, b and v as
f(a,b,v) = \delta\left(v-y(a,b)\right)f(a)f(b)
where \delta is the Dirac delta function.

The pdf f(v) of v is then the marginal of f(a,b,v) with respect to a and b
f(v) = \int\limits_{a=-\infty}^{\infty} \int\limits_{b=-\infty}^{\infty} \delta\left(v-y(a,b)\right) f(a)f(b) db da


The integral of g(x) can as stated above be evaluated using its primitive G(x):
\int\limits_{a}^{b} g(x) dx = G(b)-G(a) \quad \texttt{if} \quad b \geq a,
\int\limits_{a}^{b} g(x) dx = G(a)-G(b) \quad\texttt{if} \quad a \geq b

We thus need to account for the two possible cases:

f(v) =
\int\limits_{a=-\infty}^{\infty} \int\limits_{b=-\infty}^{a}
\delta\left(v-G(a)+G(b)\right)
f(a)f(b) db da
+\int\limits_{a=-\infty}^{\infty} \int\limits_{b=a}^{\infty}
\delta\left(v-G(b)+G(a)\right)
f(a)f(b) db da