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:

(f∗f)(x) = Z∞

−∞

f(y)f(x−y)dy

=1

π2Z∞

−∞

1

(1 + y2)

1

1 + (x−y)2dy

=1

π2

x(arctan y−arctan(x−y)) + ln 1+y2

1+(x−y)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 n∈Z≥0, show that

Γµn+1

2¶=(2n−1)!!

2n√π.

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)(2n−1)!!

2n√π

=(2n+ 1)(2n−1)!!

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

0≤x(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.

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