X and Y are independent exponential random variables each with a mean of 1. Find P(X+Y < 1)
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Can anybody explain why the answer is 1-(2e^-1) ????
I only get 1-(e^-1) based on the integral of the pdf of the convolution which is:
integral([0 to 1] of te^-t dt)
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Can anybody explain why the answer is 1-(2e^-1) ????
I only get 1-(e^-1) based on the integral of the pdf of the convolution which is:
integral([0 to 1] of te^-t dt)
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First of all, Z = X+Y has pdf
f(z) = ze^(-z) for z ≥ 0 and 0 otherwise.
Link (with derivation; see example 7.4):
http://www.dartmouth.edu/~chance/teachin…
Hence, P(X+Y < 1)
= P(Z < 1)
= ∫(z = 0 to 1) ze^(-z) dz
= ∫(w = 0 to -1) -we^w * -dw, letting w = -z
= -∫(w = -1 to 0) we^w dw
= -(we^w - e^w) {for w = -1 to 0}, via integration by parts
= 1 - 2e^(-1).
I hope this helps!
f(z) = ze^(-z) for z ≥ 0 and 0 otherwise.
Link (with derivation; see example 7.4):
http://www.dartmouth.edu/~chance/teachin…
Hence, P(X+Y < 1)
= P(Z < 1)
= ∫(z = 0 to 1) ze^(-z) dz
= ∫(w = 0 to -1) -we^w * -dw, letting w = -z
= -∫(w = -1 to 0) we^w dw
= -(we^w - e^w) {for w = -1 to 0}, via integration by parts
= 1 - 2e^(-1).
I hope this helps!