Given X~exp($ lambda $)

CDF of X = 1 - exp(-lambda*x) = P[X<=x]

P[X>x] = exp(-lambda*x)

P[Y>k] = P[X>k] = exp(-lambda*k)


P[Y>k] = P[Y=k+1] + P[Y=k+2] + ... -- (1)

P[Y>k-1] = P[Y=k] + P[Y=k+1] + P[Y=k+2] + ... -- (2)

        = P[Y=k] + P[Y>k]

then find P[Y=k] ...


Alternatively, without using the hint given, we can also approach the problem using the PDF of X

PDF of X = lambda*exp(-lambda*x)

For Y to have a value k (k is integer), X has to fall within the range of k-1 to k

P[Y=k] = P[k-1<X<k]

      = (integ:k-1 to k) lambda*exp(-lambda*x) dx = ...


Both methods should lead to the same answer.

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva