Let X And Y Be Independent Random Variables, Uniformly Distributed In The Interval [0, 1 Find The CDF
Let X and Y be independent random variables, uniformly distributed in the interval [0, 1 ]. The CDF of X – Y is FZ(z) = (1/2)(1+z)^2 for -1 ≤ z ≤ 0, 1 – (1/2)(1-z)^2 for 0 ≤ z ≤ 1, 0 for z < -1 or z > 1. The PDF of X – Y is fZ(z) = z + 1 for -1 < z < 0, 1 – z for 0 < z < 1, 0 otherwise.
To find the CDF of X – Y, we first note that the range of X – Y is [0, 1]. Let Z = X – Y, then:
FZ(z) = P(Z ≤ z) = P(X – Y ≤ z)
We can write this as an integral over the joint distribution of X and Y:
FZ(z) = ∫∫[X – Y ≤ z] fXY(x, y) dx dy
Since X and Y are independent, the joint distribution is simply the product of their marginal distributions:
fXY(x, y) = fX(x) fY(y) = 1 * 1 = 1
for 0 ≤ x, y ≤ 1.
Thus, we have:
FZ(z) = ∫∫[X – Y ≤ z] dx dy
= ∫∫[Y ≤ X – z] dx dy
= ∫0^1 ∫0^(x-z) 1 dy dx + ∫0^1 ∫(x-z)^1 1 dy dx
= ∫0^(1+z) (1-z) dx
= (1/2)(1+z)^2 for -1 ≤ z ≤ 0
= 1 – (1/2)(1-z)^2 for 0 ≤ z ≤ 1
Therefore, the CDF of X – Y is:
FZ(z) =
(1/2)(1+z)^2 for -1 ≤ z ≤ 0
1 – (1/2)(1-z)^2 for 0 ≤ z ≤ 1
0 for z < -1 or z > 1
To find the PDF of X – Y, we differentiate the CDF:
fZ(z) = dFZ(z)/dz =
z + 1 for -1 < z < 0
1 – z for 0 < z < 1
0 otherwise
Therefore, the PDF of X – Y is:
fZ(z) =
z + 1 for -1 < z < 0
1 – z for 0 < z < 1
0 otherwise
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