solutions to homework assignment 5 - indiana...
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ECE302 Spring 2015 HW5 Solutions March 27, 2015 1
Solutions to Homework Assignment 5
Problem 3.6.3 and Solution
Given X has CDF
FX(x) =
0 x < −1
0.2 x ∈ [−1, 0)0.7 x ∈ [0, 1)
1 x ≥ 1
(1)
(a) Find PV (v) where V = g(X) = |X|.
First, we find PX(x) to be
PX(x) =
0.2 x = −1 (0.2− 0)0.5 x = 0 (0.7− 0.2)0.3 x = 1 (1− 0.7)
0 otherwise
(2)
by subtracting the value of FX(x) at the lower edge of a discontinuity fromFX(x) at the upper edge of the discontinuity. Then from its definition,
SV = {0, 1}
and
PV (v) =
0.5 v = 1 (0.2 + 0.3)0.5 x = 0
0 otherwise.(3)
(b) Find FV (v).
FV (v) =
0 v < 0
0.5 0 ≤ v < 11 v ≥ 1
(4)
(c) Find E[V ].
E[V ] =∑SV
vPV (v) = 1(1/2) + 0(1/2) = 1/2. (5)
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ECE302 Spring 2015 HW5 Solutions March 27, 2015 2
Problem 3.6.4
Given: The transaction fee f for a stock purchase d is
f(d) =
{d0.01de d ≤ 10, 000
d100 + 0.005(d− 10, 000)e d > 10, 000(6)
where we are redefining the ceiling function to indicate rounding up to thenearest cent.
Consider stock with price per share s ∈ {99.75, 100, 100.25} having PMF
PS(s) =
{1/3 p ∈ {99.75, 100, 100.25}
0 otherwise.(7)
Find the PMF of the cost C of 100 shares.
Problem 3.6.4 Solution
The cost isC = d + f(d) where d = 100s.
We then calculate the fees for each possible cost of the block of 100 shares.
d = 100(99.75) = 9975 f(d) = .01(100)99.75 = 99.75
d = 100(100) = 10, 000 f(d) = .01(100)100 = 100
d = 100(100.25) = 10, 025 f(d) = d100 + 0.005(10025− 10000)e= d100.125e = 100.13
(8)
Thus
PC(c)
{1/3 c ∈ {10074.75, 10100, 10125.13}
0 otherwise.(9)
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