STA347H1F Term Test 1 (LEC 5201)
October 19th, 2021 6:10pm - 7:40pm EST
All relevant work must be shown for full marks
Total Marks = 40
Need to show all your work to receive full marks
Note: There are 4 questions (with parts) and you have to answer all of them to receive full marks.
The answers can be hand written or typed. You can submit multiple files but please make sure all
the answers are submitted in pdf, docx, jpg or png files. You will not be able to submit anything
after 7:40pm EST (other than a few exceptions). The question are created in a way that all of them
can be answered within 1 hours and 15 minutes. Please make sure that you have enough time to
scan and submit.
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Best of luck!
1
Question 1
(a) [2 Marks] Let X ∼ Geometric(θ), x = 0, 1, 2, ..... Calculate the MGF of X
Solution: The MGF,
mX(t) = E(e
tX) =
∞∑
x=0
etx(1− θ)xθ
= θ
∞∑
x=0
(
et(1− θ))x
=
θ
1− et(1− θ)
where, et(1− θ) < 1 or t < − log(1− θ)
(b) [4 Marks] Let Y ∼ Negative Binomial(r, θ). Calculate the MGF of Y . (P (Y = k) = (r+k−1r−1 )θr(1−
θ)k; k = 0, 1, 2, ...)
Solution: There are two ways to do that. The easier option is to assume Y = X1+X2+ ...Xr,
where X1, X2, ..., Xr are independent and have geometric (θ) distributions. Thus, Y = X1 +
X2 + ...Xr ∼ NB(r, θ). Thus, the MGF, mY (t) = (mX(t))r =
(
θ
1− et(1− θ)
)r
.
The second way is to use the negative-binomial formula. That is,
mY (t) = E(e
tY ) =
∞∑
y=0
ety
(
r + y − 1
r
)
θr(1− θ)y
= θr
∞∑
y=0
(
r + y − 1
r
)(
et(1− θ))y
=
θr
(1− et(1− θ))r
again, for et(1− θ) < 1 or t < − log(1− θ).
(c) [6 Marks] Using the MGF of negative binomial distribution from (b) calculate the mean and
the variance of Y .
Solution: The first derivative of the MGF,
m′Y (t) =
−r(−et)(1− θ)θr
(1− et(1− θ))r+1
m′Y (0) =
r(1− θ)θr
(θ)r+1
E(Y ) = r
(1− θ)
θ
The second derivative of the MGF,
m′′Y (t) =
ret(1− θ)θr
(1− et(1− θ))r+1 +
r(r + 1)e2t(1− θ)2θr
(1− et(1− θ))r+2
⇒ m′′Y (0) = E(Y 2) =
r(1− θ)θr
θr+1
+
r(r + 1)(1− θ)2θr
θr+2
=
r(1− θ)(1 + r − rθ)
θ2
2
Thus, the variance,
V (Y ) = E(Y 2)− (E(Y ))2
=
r(1− θ)(1 + r − rθ)
θ2
−
(
r
(1− θ)
θ
)2
=
r(1− θ) + r2(1− θ)2
θ2
− r
2(1− θ)2
θ2
=
r(1− θ)
θ2
Question 2
x
y
A
O
r
B
Z
θ
The above circle is a unit circle. That is the radius of the circle is r = 1 unit. The triangle
OAB is right triangle with the angle ∠OBA = pi2 . The angle θ is a random variable which follows
Uniform(0, pi2 ). Let, Z be the length of AB, which depends on θ.
(a) [5 Marks] Calculate E(eθ). Find its relation with eE(θ) (i.e., equal, smaller or larger).
Solution: Since θ has a Uniform(0, pi2 ) distribution, hence, E(θ) = pi4 and eE(θ) = e
pi
4 . Now,
E(eθ) =
∫ pi
2
0
eθ
1
pi
2
dθ
=
2
pi
[
eθ
]pi
2
0
=
2
pi
(
e
pi
2 − 1
)
Since, eθ is a convex function, from Jensen’s inequality we can see that eE(θ) < E(eθ). Since,
we know that it is not a linear function.
Note: Or one can just calculate the values that E(eθ) = 2.425825 and eE(θ) = 2.19328. Hence,
eE(θ) < E(eθ)
(b) [5 Marks] Show that E(Z) < 1√
2
.
Solution: We need to identify that Z = sin(θ). Since sin(θ) is a concave function between 0
to pi/2, thus, E(Z) = E(sin(θ)) ≤ sin(E(θ)) = sin(pi/4) = 1√
2
, by using Jensen’s inequality.
3
Question 3
A man is given n different keys, only one of which will open his door. He tries them successively
until he can open the door and then he stops.
(a) [3 Marks] Describe the sample space of this experiment.
Solution: There are n keys in total. The possibilities are S, FS, FFS, FFFS, ...,, where the
sequence ends on the nth try. Let’s define
Si =
{
1, if the door opens at the ith try
0, if the door does not open at the ith try
Thus, the sample space = {S1, S2, ..., Sk|k ≤ n, Sk = 1}
(b) [5 Marks] Show that each outcome of the experiment has the same probability. What is the
probability?
Solution: The probability that he will be successful at the first try, P (S) = 1n . The probability
that he will be successful at the second try is P (FS) = n−1n 1n−1 = 1n . The probability of success
on the third try P (FFS) = n−1n n−2n−1 1n−2 = 1n . As we can see that ∀k ≤ n, P (Sk) = 1n .
Question 4
(a) [3 Marks] Consider, for α > 0 fixed, a function is given by f(x) = α(1+x)−α−1, for 0 < x <∞.
Show that f(.) is a density.
Solution: We have that f(x) ≥ 0 for every x. We have,∫ ∞
0
α(1 + x)−α−1dx =
[
− (1 + x)−α
]∞
0
= 1
Thus, f(x) is a density.
(b) [3 Marks] Calculate the CDF of the distribution provided in (a).
Solution: The CDF can be given as,
F (x) =
∫ x
0
α(1 + u)−(α+1)dz = 1− (1 + x)−α (1)
(c) [3 Marks] Calculate the median of the distribution given in (a).
Solution: From the CDF we can calculate,
1− (1 +M)−α = 1
2
⇒(1 +M)−α = 1
2
⇒1 +M = 2 1α
⇒M = 2 1α − 1
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