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14.30 Introduction to Statistical Methods in Economics
Spring 2009
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View Full Document Problem Set
#5

Solutions
14.30

Intro. to Statistical Methods in Economics
Instructor: Konrad Menzel
Due: Tuesday, March 31, 2009
Question
One
The convolution formula is a useful trick when we are interested in the sum or average of
independent random variables. In the last problem set, we dealt with the random variable
X, below.
e"
forx
>
0
f
(4
=
0
forx<O
Now, suppose that
X
=
X1
=
X2
=
.
=
XI, are independent, identically distributed
random variables.
1. Using the convolution formula, determine the PDF of
&
=
(xl
+
X2). Hint: Defilze
Zl
=
X1 and Z2
=
X1
+
X2 and then use the transformation method to get back to Y2
from Z2.
Solution to 1: To use the convolution formula, we need the joint PDF of
X1 and
X2 and x2 as a function of y2 and xl. The function is x2
=
2y2

xl. Also, since
they are independent, we can just construct the joint PDF by multiplying the two
marginals, fxl(xl) and fx2(x2). This gives a joint PDF of
f (xl, x2)
=
e(x1+x2)
for xl
>
0 and x2
>
0 and zero otherwise. The convolution formula adapted to
this problem (taking into account the limits) is (where we take into account the
change in variables formula for x2 as a function of xl and y2 using the Jacobian)
2. Compute its expectation: IE[Y2].
Solution to 2: To compute the expectation, we use the standard formula for
continuous random variables:
This integral must be computed using two iterations of integration by parts.
Alternatively, we could have used the properties of expectations (no convolution
necessary) to obtain
3.
Using the convolution formula, determine the
PDF for
&
=
;(XI
+
X2
+
X3).
Hint:
Use the hint from part 1 to define
Z3
=
XI
+
X2
+
X3
and perform a convolution with
and
Z2
to transform the problem into
Z3.
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View Full Document Solution to 3: Apply the same method to the results from part
I, where we
obtained fY2(y2).We write the joint PDF of
XI, X2,and X3 as
We'll take a slightly different approach for this problem, in order to help us solve
part
5 later. Define Z2
=
XI
+X2
and Z3
=
Xl +X2 +X3. Now, we can ignore any
Jacobian transformation and directly apply the convolution formula. We can first
solve for 22's distribution by only using the convolution on the joint distribution
of
XI and X2.
Now, we use the convolution on the joint distribution of Z2 and
X3 to obtain the
distribution of Z3:
But, we were interested in Y3, not Z3. SO,we perform a simple continuous inverse
transformation of random variables using Z3
=
3Y3:
4. Compute its expectation: IE[Y3].
Solution to 4: Very easily we can apply the properties of expectations to determine
again that IE[Y3]
=
1. However, we can also determine this using the convolution
formula. What we should have learned from part 2 was that all of the leading
integration by parts terms will not matter as
yGey
=
0 for y
=
oo
and for y
=
0
for any
k
E
N.
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This note was uploaded on 01/30/2010 for the course STAT 430 taught by Professor Jones during the Fall '10 term at Napa Valley College.
 Fall '10
 jones

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