Solutions Assignment 4 - Probability and Phase Transition | STAT 330, Assignments of Statistics

Material Type: Assignment; Class: PROBAB&STAT COM SCI; Subject: STATISTICS; University: Iowa State University; Term: Unknown 1992;

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Stat 330 Solution to Homework 4
1 Probability Mass Function
Let Xbe 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:
x0 1 2 3
pX(x) 0.5 0.25 0.1
Since the sum of the probabilities has to be 1 for a probability mass function, pX(3) = 1 0.50.25
0.1=0.15
(b) Derive the cumulative distribution function for Xand draw it in a chart.
x0 1 2 3
FX(x) 0.5 0.75 0.85 1
(c) Determine the probabilities that...
(a) Xis at least 2.
P(X2) = 0.1+0.15 = 0.25
(b) Xis neither 0 nor 2.
P(X6= 0,2) = 1 P(X= 0) P(X= 2) = 1 0.50.1=0.4
(c) Xis non-negative.
P(X0) = 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[X2]=0·P(X= 0) + 1 ·P(X= 1) + 4 ·P(X= 2) + 9 ·P(X= 3) = 0 + 0.25 + 0.4+1.35 = 2
V ar[X] = E[X2](E[X])2= 2 0.92= 1.19
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Stat 330 Solution to Homework 4

1 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 0 1 2 3 pX (x) 0. 5 0. 25 0. 1

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 0 1 2 3 FX (x) 0. 5 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

V ar[X] = E[X^2 ] − (E[X])^2 = 2 − 0. 92 = 1. 19

(e) Let Y be a random variable with Y = 5 − 2 X. 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 − 2 X] = 5 − 2 E[X] = 5 − 1 .8 = 3. 2

V ar[Y ] = V ar[5 − 2 X] = V ar[− 2 X] = (−2)^2 V ar[X] = 4 · 1 .19 = 2. 38

2 Calculus with Variances

(a) Use the definition of variance to show that V ar(aX) = a^2 V ar(X) for any value a ∈ R

V ar(aX) = E

[

(aX − E[aX])^2

]

= E

[

(aX − aE[X])^2

]

= E

[

a^2 (X − E[X])^2

]

= a^2 E

[

(X − E[X])^2

]

= a^2 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 )

3 Market Shares, Baron p.

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 X -3 0 3 pX 0.3 0.2 0.

Y -3 0 3

pY 0.4 0 0.

(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: