5. Suppose X and Y are two independent Uniform(0, 1) random variables. Use the cumulative distribution function method to find the probability density function of their sum U = X+Y. Solution: Given that X = (0,1) and Y = (0, 1), we have U € (0,2). Then for u € (0,2), the equality U = u implies that X = u-Y. Hence, since we want 0 Uu, we observe that we need 0≤x≤u-Y, which is possible if and only if Y

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 22E
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Could you explain how the inequalities u in (0,1), we have 0 ≤ X ≤u-Y for any 0 ≤Y<u and u in (1,2), we either have 0 ≤ X ≤u-Y for any u - 1 < Y<1, or 0≤x≤1 for any 0 ≤Y≤u - 1 are obtained please. They're in the solutions but don't understand how they were derived.

5. Suppose X and Y are two independent Uniform(0, 1) random
variables. Use the cumulative distribution function method
to find the probability density function of their sum U = X+Y.
Transcribed Image Text:5. Suppose X and Y are two independent Uniform(0, 1) random variables. Use the cumulative distribution function method to find the probability density function of their sum U = X+Y.
Solution: Given that X = (0,1) and Y = (0, 1), we have U €
(0,2).
Then for u € (0,2), the equality U = u implies that X = u-Y.
Hence, since we want 0 Uu, we observe that we need
0≤x≤u-Y, which is possible if and only if Y<u.
However, given that 0 < x < 1 and 0 < y < 1, we need to split
into two cases:
For u € (0,1), we have 0 ≤ X ≤u-Y for any 0 ≤Y<u, thus
Fu (u) = P(U≤u) = P(X + Y≤u)
= P(X ≤u-Y)
26 ru-y
= "fx.x(x, y) drdy
CH
=
u-y
dxdy
= L" (u− y) dy
1
=
= u²
y-0
For u (1,2), we either have 0 ≤ X ≤u-Y for any u - 1 <
Y<1, or 0≤x≤1 for any 0 ≤Y≤u - 1, thus
Fu(u) = P(U≤u) = P(X + Y≤u)
= P(X ≤u-Y)
=fxx(x, y) dxdy +
=LC
pu-y
=
=
cu-1
ddy+ T
(u - y) dy +
1
Cu-1
fx,y(x, y) dxdy
dxdy
cu-1
dy
Jo
1
+u-12u-
-1.
So, we have
y-u-1
fu (u) =
=
dF(u)
du
-E-
=
if 0<u<1,
2-u, if 1 < u<2,
0
otherwise.
Transcribed Image Text:Solution: Given that X = (0,1) and Y = (0, 1), we have U € (0,2). Then for u € (0,2), the equality U = u implies that X = u-Y. Hence, since we want 0 Uu, we observe that we need 0≤x≤u-Y, which is possible if and only if Y<u. However, given that 0 < x < 1 and 0 < y < 1, we need to split into two cases: For u € (0,1), we have 0 ≤ X ≤u-Y for any 0 ≤Y<u, thus Fu (u) = P(U≤u) = P(X + Y≤u) = P(X ≤u-Y) 26 ru-y = "fx.x(x, y) drdy CH = u-y dxdy = L" (u− y) dy 1 = = u² y-0 For u (1,2), we either have 0 ≤ X ≤u-Y for any u - 1 < Y<1, or 0≤x≤1 for any 0 ≤Y≤u - 1, thus Fu(u) = P(U≤u) = P(X + Y≤u) = P(X ≤u-Y) =fxx(x, y) dxdy + =LC pu-y = = cu-1 ddy+ T (u - y) dy + 1 Cu-1 fx,y(x, y) dxdy dxdy cu-1 dy Jo 1 +u-12u- -1. So, we have y-u-1 fu (u) = = dF(u) du -E- = if 0<u<1, 2-u, if 1 < u<2, 0 otherwise.
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