Quote:
Originally Posted by Justin A
Let X and Y be independent exponential RVs with rates lambda and mu, lambda > mu.
How do I find the conditional density function of X, given that X + Y = c, where c is some constant > 0?
If X and Z have joint density f(x,z) and marginal densities f_X(x), f_Z(z), then the conditional density of X, given Z = z, is f(x|z) = f(x,z)/f_Z(z). Therefore, we have
f(x|z) = f(z|x)f_X(x)/f_Z(z).
In this case, if Z = X + Y, then the conditional distribution of Z given X is easy to calculate:
P(Z ≤ z | X = x) = P(Y ≤ z - x) = 1 - e^{-
μ(z - x)}.
Differentiating gives
f(z|x) =
μe^{-
μ(z - x)}.
Hence,
f(x|z) =
μe^{-
μ(z - x)}
λe^{-
λx}/f_Z(z).
In other words, for fixed c, f(x|c) is proportional to e^{-(
λ -
μ)x}. Since the integral of f(x|c) over [0,c] must be 1, it is easy to calculate the constant of proportionality (which depends on c,
λ, and
μ).
