Random Variables - Probability - Solved Exam, Exams for Probability. Punjab Engineering College

Probability

Description: This is the Solved Exam of Probability which includes Exponentially Distributed, Continuous, Random Variable, Expected, Positive Integer, Geometric Distribution, Probability, Geometrically, Distributed, Value etc. Key important points are: Random Variables, First Line, Amount, Assignment, Cauchy Distributed, Independent, Density Function, Partial Fractions, Convolution Product, Logarithm Approach
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Probability: Problem Set 6
Fall 2009
Instructor: W. D. Gillam
Due Nov. 6, start of class
Instructions. Print your name in the upper right corner of the paper and write “Problem
Set 6” on the first line on the left. Skip a few lines. When you finish this, indicate on
the second line on the left the amount of time you spent on this assignment and rate its
difficulty on a scale of 1 5 (1 = easy, 5 = hard).
(1) Let X, Y be Cauchy distributed independent random variables. Calculate the
density function for X+Y.
Solution. The density function for the Cauchy distribution is
f(x) = 1
π(1 + x2)
so the density for X+Yis given by the convolution product, which can be inte-
grated by partial fractions:
(ff)(x) = Z
−∞
f(y)f(xy)dy
=1
π2Z
−∞
1
(1 + y2)
1
1 + (xy)2dy
=1
π2
x(arctan yarctan(xy)) + ln 1+y2
1+(xy)2
4x+x3
−∞
=2
π(4 + x2)
= (1/2)f(x/2).
Note that the terms in the logarithm approach 1 as y→ ±∞, so they drop out
and we have used
lim
y→±∞ arctan y=±π
2.
Notice that the density for X+Yis the same as the density for 2Xin this case.
This is not generally true, as we know from the Central Limit Theorem; the Cauchy
distribution shows the necessity of the assumptions of finite expected value and
variance (or at least some assumptions along these lines...) in the Central Limit
Theorem. For example, even when Xhas a uniform distribution, X+Ywill have
a Simpson distribution even though 2Xalso has a uniform distribution.
(2) For nZ0, show that
Γµn+1
2=(2n1)!!
2nπ.
Hint: When n= 0, this is the Gaussian integral (note (1)!! = 1 by convention).
Induct on nand use the functional equation Γ(z+ 1) = zΓ(z).
1
2
Solution. By induction, we compute, using the functional equation:
Γ(n+ 1 + 1
2) = (n+1
2)Γ(n+1
2)
=(n+1
2)(2n1)!!
2nπ
=(2n+ 1)(2n1)!!
2n+1 π
=(2n+ 1)!!
2n+1 π.
(3) Let Xbe a continuous random variable with cumulative distribution function F(x)
and density function f(x). Assume
Z
−∞ |x|f(x)dx
exists.1Show that
lim
x+x(1 F(x)) = 0
lim
x→−∞ xF (x) = 0.
Show that
E(X) := Z
−∞
xf(x)dx
=Z
0
(1 F(x))dx Z0
−∞
F(x)dx.
Draw a typical such Fand shade the regions whose areas are given by these
integrals. You may want to think about the analog of this when the expected
value is replaced by the variance.
Solution. Since Z
−∞ |x|f(x)dx
exists, we must have:
lim
x→∞ Z
x
yf(y)dy = 0
lim
x→−∞ Zx
−∞ yf(y)dy = 0.
For positive x, we have
0x(1 F(x)) = xZ
x
f(y)dy
Z
x
yf(y)dy.
1Technically speaking, one should assume this exists to even speak of the expected value, though I have not
really emphasized this much in class except when this arose in connection with the Cauchy distribution.
3
Taking the limit x and applying the Squeeze Theorem to the above inequality
yields
lim
x→∞ x(1 F(x)) = 0.
Similarly, for x < 0, we have
0xF (x) = xZx
−∞
f(y)dy
Zx
−∞
yf(y)dy.
Taking the limit x→ −∞ and applying the Squeeze Theorem to this inequality
yields
lim
x→−∞ xF (x) = 0
(the minus sign is no big deal since minus signs commute with limits and 0 = 0).
For the second statement, use integration by parts to write:
Zx
0
yf(y)dy =Zx
0
ydF (y)
= [yF (y)]x
0Zx
0
F(y)dy
=xF (x)Zx
0
F(y)dy
=x(1 F(x)) + Zx
0
(1 F(y))dy
and similarly:
Z0
x
yf(y)dy =Z0
x
ydF (y)
= [yF (y)]0
xZ0
x
F(y)dy
=xF (x)Z0
x
F(y)dy.
Now add these two and take the limit x→ ∞. Two of the four terms drop out by
the previous part.
(4) Let f(x) be the density function for...
(a) the standard normal distribution.
(b) the gamma distribution with parameters α, β > 0.
(c) the beta distribution with parameters α, β > 0.
(d) Student’s distribution.
Show that there are constants a, b, c, d R(depending on the parameters α, β for
the beta and gamma distributions; you should calculate these constants in each
case) such that
d
dx ln f(x) = x+a
b+cx +dx2.
4
Solution. We’ll just do the standard normal f(x) = 1
2pi ex2/2to give the idea.
d
dx ln f(x) = f(x)
f(x)
=x.
This is of the above form with a=c=d= 0 and b=1.
(5) Let X, Y be independent random variables whose expected values and variances
exist. Prove that
V(XY ) = V(X)V(Y) + E(X)2V(Y) + E(Y)2V(X).
Solution. This is a straightforward computation with moments.
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