Probability 2017 Homework 3
March 20, 2017
1
Q1 (10 points)
Suppose that X is nonnegative continuous random variable. Prove that E[X] =
R +∞
0
P (X > x)dx.
Solution:
Suppose f (x) is the density function of X.
Z
+∞
Z
+∞
Z
+∞
f (y)dydx
P (X > x)dx =
0
0
Z
Z
=
0
Z
=
f (y)dy)
x
x
y
+∞ Z
f (y)dxdy
Z y
+∞
f (y)
dxdy
(by changing the order of integral)
0
0
Z
+∞
(by P (X > x) =
0
(1)
+∞
=
f (y)ydy
0
Z
+∞
=
f (y)ydy
(by the nonnegtivity of X)
−∞
= E[X]
2
Q2 (10 points)
Consider a triangle and a point chosen within the triangle according to the uniform probability
law. Let X be the distance from the point to the base of the triangle. Given the height of the
triangle, find the CDF and the PDF of X.
Solution:
Suppose 4ABC is the considered triangle. Let h be the height of 4ABC, S be the area of 4ABC
. Draw a line parallel to the base within the triangle, whose distance to the base is x. The intersect
of this line and the waist of the triangle is D and E respectively.
1
Then for 0 < x < h:
P (X > x) = P (the point f alls in 4ADE)
= S4ADE /S
2
(by unif orm probability law)
2
= (h − x) /h
(2)
(by the property of similar triangles)
Thus the CDF of X is:
P (X ≤ x) = 1 − P (X > x)

2
2

1 − (h − x) /h , 0 < x < h
= 0,
x≤0


1,
x≥h
(3)
The PDF is obtained by differentiating CDF:
d
[P (X ≤ x)]
dx
(
2(h − x)/h2 , 0 < x < h
=
0,
otherwise
f (x) =
3
(4)
Q3 (10 points)
Please write down the PDF of a normal random variable with mean µ and standard deviation σ.
Solution:
Suppose X is a normal random variable with mean µ and standard deviation σ. Then the PDF of
X is:
(x−µ)2
1
(x ∈ IR)
e− 2σ2
f (x) = √
2πσ
.
4
Q4 (10 points)
A point is chosen at random (according to a uniform PDF) within a semicircle of the form {(x, y) :
x2 + y 2 < r2 , y > 0}, for some given r>0.
a) Find the joint PDF of the coordinates X and Y of the chosen point.
b) Find the marginal PDF of Y and use it to find E[Y ].
Solution:
a) Let B = {(x, y) : x2 + y 2 < r2 , y > 0}. the joint PDF of the coordinates X and Y is:
(
2/πr2 , (x, y) ∈ B
f (x, y) =
0,
otherwise
(5)
b) The marginal density of Y on (0, r) is:
Z
+∞
fY (y) =
f (x, y)dx
−∞
√
Z
+
=
√
−
=
r 2 −y 2
r 2 −y 2
2
dx
πr2
(6)
4 p 2
r − y2
πr2
Thus, the marginal density of Y is:
(
fY (y) =
4
πr 2
p
r2 − y2 ,
0,
2
y ∈ (0, r)
otherwise
(7)
The expectation of Y is:
Z
+∞
E[Y ] =
yfY (y)dy
−∞
Z r
=
y
0
=
5
4 p 2
r − y 2 dy
πr2
(8)
4r
3π
Q5 (10 points)
Let X1 , X2 , ..., Xn be independent random variables. We know that variance has linearity for
independent random variables, namely
V ar[X1 + X2 + ... + Xn ] = V ar[X1 ] + V ar[X2 ] + ... + V ar[Xn ]
How about the variance of the product? Can you express
V ar[X1 X2 ...Xn ]
in terms of V ar[Xi ] and E[Xi2 ] ?
Solution:
First note two facts:
Fact 1: if X1 , X2 , ..., Xn are independent, then X12 , X22 , ..., Xn2 are also independent.
Fact 2: if X and Y are independent, then E[XY ] = E[X]E[Y ].
Then,
V ar[X1 X2 ...Xn ] ≡ E[(X1 X2 ...Xn − E[X1 X2 ...Xn ])2 ]
= E[(X1 X2 ...Xn )2 − 2E[X1 X2 ...Xn ](X1 X2 ...Xn ) + E(X1 X2 ...Xn )2 ]
= E[(X1 X2 ...Xn )2 ] − 2E[X1 X2 ...Xn ]2 + E[X1 X2 ...Xn ]2
= E[(X1 X2 ...Xn )2 ] − E[X1 X2 ...Xn ]2
= E[(X12 X22 ...Xn2 )] − E[X1 X2 ...Xn ]2
= E[X12 ]E[X22 ]...E[Xn2 ] − E[X1 X2 ...Xn ]2
=
=
=
(by F act 1 and F act 2)
E[X12 ]E[X22 ]...E[Xn2 ] − E[X1 ]2 E[X2 ]2 ...E[Xn ]2 (by F act 2)
E[X12 ]E[X22 ]...E[Xn2 ] − [E[X12 ] − V ar(X1 )][E[X22 ] − V ar(X2 )]...[E[Xn2 ]
n
n
Y
Y
E[Xi2 ] −
[E[Xi2 ] − V ar(Xi )]
i=1
i=1
− V ar(Xn )]
(9)
3