STAT 510 HW #4 – Answers
Due Friday, September 29, 2006
You can turn it in in class, or to my office (116B IH) or mailbox in 101 IH, by 4PM.
There are also questions on Mallard.
1. The double exponential random variable Xhas space Rand pdf f(x) = (1/2)e−|x|.
(a) Show that the mgf of a double exponential is MX(t) = 1/(1 −t2). [Break the integral
into two parts according to the sign of x.] For which tis it finite?
Answer:
MX(t) = E[etX ] = 1
2Z∞
−∞ etx−|x|dx =1
2Z0
−∞ etx+xdx +1
2Z∞
0etx−xdx.
The first integral is
Z0
−∞ etx+xdx =Z0
−∞ ex(1+t)dx =1
1 + tex(1+t)|0
−∞ =1
1 + tif 1 + t > 0.
The second is Z∞
0etx−xdx =Z∞
0e−x(1−t)dx =1
1−tif 1−t > 0,
as for the exponential. So the mgf is finite if both 1 + t > 0 and 1 −t > 0, i.e., if −1< t < 1.
If so, then
MX(t) = 1
21
1−t+1
1 + t=1
2
1 + t+ 1 −t
(1 −t)(1 + t)=1
1−t2.
(b) Suppose Uand Vare independent Exponential(1), and let Y=U−V. Find the
mgf of Y. What is the distribution of Y?
Answer: The mgf of an E xponential(1) is 1/(1 −t) if t < 1, hence
MY(t) = E[etY ] = E[et(U−V)] = E[etU ]E[e−tV ] = 1
1−t
1
1 + t=1
1−t2
when t < 1 and −t < 1, i.e., when −1< t < 1. So Yis double exponential.
(c) The mean of a double exponential is 0 and the variance is 2. Suppose X1, . . . , Xnare
independent and identically distributed double exponentials, and let Wnbe the standardized
mean,
Wn=X
q2/n.
Show that the mgf of Wnis
Mn(t) = 1
(1 −t2
2n)n.
1
