Stat 330
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Solution to Homework 4
Probability Mass Function
Let X be a random variable with image Im(X) = {0, 1, 2, 3}.
(a) Fill in the blank in the table below to make it a valid probability mass function:
x
pX (x)
0
0.5
1
2
0.25 0.1
3
Since the sum of the probabilities has to be 1 for a probability mass function, pX (3) = 1 − 0.5 − 0.25 −
0.1 = 0.15
(b) Derive the cumulative distribution function for X and draw it in a chart.
x
FX (x)
0
0.5
1
2 3
0.75 0.85 1
(c) Determine the probabilities that...
(a) X is at least 2.
P (X ≥ 2) = 0.1 + 0.15 = 0.25
(b) X is neither 0 nor 2.
P (X 6= 0, 2) = 1 − P (X = 0) − P (X = 2) = 1 − 0.5 − 0.1 = 0.4
(c) X is non-negative.
P (X ≥ 0) = 1
(d) Find the expected value and variance of X.
E[X] = 0 · P (X = 0) + 1 · P (X = 1) + 2 · P (X = 2) + 3 · P (X = 3) = 0 + 0.25 + 0.2 + 0.45 = 0.9
E[X 2 ] = 0 · P (X = 0) + 1 · P (X = 1) + 4 · P (X = 2) + 9 · P (X = 3) = 0 + 0.25 + 0.4 + 1.35 = 2
2
V ar[X] = E[X 2 ] − (E[X]) = 2 − 0.92 = 1.19
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(e) Let Y be a random variable with Y = 5 − 2X. Determine the image of Y . Based on the rules for
expected values and variances, find the expected value and variance of Y .
Since X has image im(X) = {0, 1, 2, 3}, the image of Y has to be im(Y ) = {5, 3, 1, −1}
and
E[Y ] = E[5 − 2X] = 5 − 2E[X] = 5 − 1.8 = 3.2
V ar[Y ] = V ar[5 − 2X] = V ar[−2X] = (−2)2 V ar[X] = 4 · 1.19 = 2.38
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Calculus with Variances
(a) Use the definition of variance to show that V ar(aX) = a2 V ar(X) for any value a ∈ R
V ar(aX)
= E (aX − E[aX])2 =
= E (aX − aE[X])2 =
= E a2 (X − E[X])2 =
= a2 E (X − E[X])2 = a2 V ar(X)
(b) Show that V ar(X − Y ) = V ar(X) + V ar(Y ) for two independent random variables X and Y . Use part
(a) and the property that V ar(X + Y ) = V ar(X) + V ar(Y ) for two independent random variables X
and Y .
V ar(X − Y )
= V ar(X + (−1)Y ) =
= V ar(X) + V ar(−Y ) = V ar(X) + (−1)2 V ar(Y ) = V ar(X) + V ar(Y )
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Market Shares, Baron p.67
Shares of company A cost $10 per share and give a profit of X%. Independently of A, shares of company B
cost $50 a share and give a profit of Y %. Deciding how to invest $1,000, you decide between three portfolios:
(a) 100 shares of A,
(b) 50 shares of A and 10 shares of B,
(c) 20 shares of B.
The probability mass functions of X and Y are given as
-3
0
3
X
pX
0.3 0.2 0.5
Y
pY
-3
0.4
0
0
3
0.6
(a) Compute expected value and variance of the total dollar profit generated by each portfolio.
Instead of looking at % return per dollar, we can more conveniently, look at dollar return for each of
the companies. Let A be the random variable for the dollar return for each share in company A and
let B be the dollar return for each share of company B, then we get:
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A -$0.3
$0 $0.3
B -$1.5 $0 $1.5
Then we can write the return in dollars for
pA
0.3 0.2
0.5
pB
0.4
0
0.6
each of the three portfolios, P1 , P2 , and P3 as linear combinations of random variables A and B:
P1
=
100A
P2
=
50A + 10B
P3
=
20B
For expected values and variances on the dollar return for each of the portfolios, we need to know
expected values and variances for random variables A and B:
E[A]
V ar[A]
=
−$0.3 · 0.3 + $0 · 0.2 + $0.3 · 0.5 = 0.06
2
= E[A2 ] − (E[A]) =
= (−$0.3)2 · 0.3 + $02 · 0.2 + ($0.3)2 · 0.5 − 0.062 = 0.0684
E[B]
=
−$1.5 · 0.4 + $1.5 · 0.6 = 0.3
V ar[B]
=
(−$1.5)2 · 0.4 + ($1.5)2 · 0.6 − 0.32 = 2.16
This gives an expected value of $6 for each of the portfolios, but different variances of
V ar(P1 )
=
1002 V ar(A) = 684
V ar(P2 )
=
502 V ar(A) + 102 V ar(B) = 387
V ar(P3 )
=
202 V ar(B) = 864
(b) If our goal is to maximize the expectation and minimize the variance, what is the least risky portfolio?
What is the most risky portfolio?
The most risky portfolio is P3 - it has the highest variance, i.e. we have more spread in the actual
dollar return, indicating that we could make a lot more money than the expected $6, but we could also
lose the most money. The mixture of shares from companies A and B make up the least risky portfolio,
which still has a variance of 387, indicating that we could lose money very easily with this portfolio.
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