Stat 430 HW 5
1. Let X be a random variable with PDF
(
c(sin2 (x) + log(x))
fX (x) =
0
if 1 < x ≤ 2
otherwise
(a) Find c
(b) Find the CDF FX (x)
(c) Find E(X) and V ar(X)
Solution
R2
(a) 1 c(sin2 (x) + log(x))dx = 1
2
c[− x2 + xlog(x) − 14 sin(2x) ] = 1
1
c ≈ 0.768
Rx
(b) FX (x) = 1 0.768(sin2 (x) + log(x))dx x
FX (x) = 0.768[− x2 + xlog(x) − 14 sin(2x) ]
1
R2
(c) E(X) = 1 0.768x(sin2 (x) + log(x))dx ≈ 1.553
R2
E(X 2 ) = 1 0.768x2 (sin2 (x) + log(x))dx ≈ 2.485
V ar(X) = E(X 2 ) − (E(X))2 = 1.5532 − 2.485 ≈ 0.073
2. You are allowed to take a certain test three times, and your final score will be the maximum of the
test scores. Your score in test i, where i = 1, 2, 3, takes one of the values from i to 10 with equal
probability 1/(11 − i), independently of the scores in other tests. What is the PMF of the final score?
Solution By independence, we have P (max {X1 , X2 , X3 } ≤ x) = P (X1 ≤ x)P (X2 ≤ x)P (X3 ≤
x−1 x
x) = x−2
8
9 10 where x ∈ 3, 4, . . . , 10. Notice that we can’t have x < 3 for the maximum because
the minimum of the third exam is 3. To compute the PMF at x, just take P (max {X1 , X2 , X3 } ≤
x) − P (max {X1 , X2 , X3 } ≤ x − 1).
3. (a) The median of a random variable X is a number µ that satisfies FX (µ) = 1/2. Find the median
of the exponential random variable with parameter λ.
(b) When is the median of a random variable not unique? If a continuous random variable has more
than one median, can it have a finite number?
(c) The mode of a discrete random variable is the set of any values that are most likely (ie, no
other value has a higher probability). For a continuous random variable, this corresponds to
0 (x) = 0 and f 00 (x) < 0. Find the mode of X where X ∼ N (0, 1) and of eX .
fX
X
(d) Find the median of eX
1
Solution
(a) The CDF of an exponential random variable is FX (x) = 1 − e−λx
So to find the median,
FX (µ) = 1 − e−λµ = 1/2
µ = ln2
λ
(b) The median is not unique when there is an interval of values where PDF is zero or CDF has
zero slope. In this case, we have uncountably infinite medians.
(c) For X ∼ N (0, 1), the mode is when X = 0. The density of eX is
√1
e− {log(x)
x 2πσ
− µ}2 /(2σ 2 ).
2
Differentiating we get the mode is eµ−σ . In this case, µ = 0, σ = 1 so the mode is e−1
(d) The median is eµ since P (X ≤ µ) = P (eX ≤ eµ ) by monotonicity of the logarithm.
4. Let X be normal with mean 1 and variance 4. Let Y = 2X + 3.
(a) What is the PDF of Y ?
(b) Find P (Y ≥ 0)
Solution
(a) X ∼ N (1, 4) → X = 1 + 2Z
Y = 2X + 3 → Y = 2(1 + 2Z) + 3 = 5 + 4Z
Therefore, Y is also a normal random variable with parameters
E(Y ) = 5
V ar(Y ) = σ 2 = 16
And,
fY (y) =
√1 e−
4 2π
(y−5)2
32
(b) P (Y ≥ 0) = 1 − P (Y ≤ 0) = 1 − Φ(− 54 ) = 0.8944
5. An old modem can take anywhere from 0 to 30 seconds to establish a connection, with all times
between 0 and 30 being equally likely.
(a) What is the probability that if you use this modem you will have to wait more than 15 seconds
to connect?
(b) Given that you have already waited 10 seconds, what is the probability of having to wait at
least 10 more seconds?
Solution
(a) For a uniform random variable T with bounds of a and b, the CDF is FT (T ≤ t) =
P (t > 15) = 1 − FT (t ≤ 15) = 1 − 15−0
30−0 = 1/2
(b) P (t > 20|t > 10) =
P (t>20,t>10)
P (t>10
=
P (t>20)
P (t>10)
=
1−20/30
1−10/30
x−a
b−a
= 1/2
6. A consulting project has two phases to it. Phase one is the analysis phase and Phase two is the
presentation and write up phase. Let X be the time required to do Phase one in months and Y be
the total time in months it takes to complete Phase two. Suppose that (X, Y ) has a joint density
f (x, y) = c where 0 ≤ x2 ≤ y ≤ 1 and otherwise f (x, y) = 0.
(a) Find c.
2
(b) What is the density function for X?
(c) If it takes 21 month to complete Phase one, what is the probability that the total time it takes
to complete Phase two is more than 43 of a month?
Solution
(a) Graphically, the region where the pdf is nonzero looks like.
1
y
x
1
Figure 1: Domain of the pdf (in blue)
Since the area of the square is 1, the area of the domain is 1 −
R1
0
x2 dx = 2/3. Thus, c = 3/2
(b) To find the marginal density of X we take
Z
1
fX (x) =
fX,Y (x, y)dy
x2
3
= (1 − x2 )
2
for x ∈ [0, 1], and 0 otherwise.
(c)
Z
1
P (Y > 3/4|X = 1/2) =
fY |X=1/2 (y)dy
3/4
Z 1
=
3/4
1
Z
=
3/4
fX,Y (1/2, y)
dy
fX (1/2)
3/2
dy
9/8
= 1/3
Another way is to look at this graphically.
1
y
x
1
Figure 2: Vertical line is where we restrict y to be.
Since we have fX , Y (x, y) is uniform, we know that in the vertical slice, Y is uniform as well.
1
And the normalizing constant is 3/4
because the line is of length 3/4.
3
7. Suppose that the quality of an item X is uniformly distributed on the interval [0, 1] and that the
profit Y is given by Y = X 5 . Find the covariance between X and Y .
Solution Cov(X, Y ) = Cov(X, X 5 ) = E(X 6 ) − E(X)E(X 5 )
For any n R> −1
1
1
E(X n ) = 0 xn dx = n+1
So the covariance is
1
1
1
5
7 − 2 × 6 = 84
8. Let the random variables X, Y have a joint density f (x, y) = cx2 y where 0 ≤ 2y ≤ x ≤ 1 and
otherwise f (x, y) = 0.
(a) Find c.
(b) Find the marginal density of X and the conditional density of Y given that X = x.
(c) Find E(Y |X = x).
Solution
R 1 R x/2
R1
(a) 0 0 cx2 ydydx = 8c 0 x4 dx =
So c = 40
R x/2
(b) fX (x) = 0 40x2 ydy = 5x4
fY |X (y) =
fX,Y (x,y)
fx (x)
(c) E(Y |X = x) =
=
R x/2
0
40x2 y
5x4
c
40
= 8 xy2
yfY |X (y)dy =
R x/2
0
2
8 xy 2 dy = 13 x
4