Download Solution Series 8 - D-MATH

Probability and statistics
May 3, 2016
Lecturer: Prof. Dr. Mete SONER
Coordinator: Yilin WANG
Solution Series 8
Q1. If log(X) has the normal distribution with mean µ and variance σ 2 , we say that X has the
lognormal distribution with parameters µ and σ 2 .
A popular model for the change in the price of a stock over a period of time of length u is to
say that the price after time u is Su = S0 Zu , where Zu has the lognormal distribution with
parameter µu and σ 2 u. In this formula, S0 is the present price of the stock, and σ is called
the volatility of the stock price.
(a) What is the expected value of S1 ?
(b) Find the distribution of 1/S1 .
(c) What is the expected value of 1/S1 ?
(d) What are k-th moments of S1 , for k = 1, 2, . . .?
Solution:
(a) By definition of a lognormal distribution, N := log(Z1 ) ∼ N(µ, σ 2 ). Thus
E(Z1 ) = E(eN ) = ϕN (1) = eµ+σ
2 /2
,
where we have used the moment generating function of N :
2 σ 2 /2
ϕN (t) = E(etN ) = etµ+t
.
We get:
E(S1 ) = S0 eµ+σ
2 /2
.
(b) Let X := (log(Z1 ) − µ)/σ ∼ N(0, 1),
µ+σX
S1 = S0 e
e−µ−σX
1
=
.
⇒
S1
S0
Hence 1/S1 has the distribution of 1/S0 times a lognormal random variable with parameter −µ and σ 2 .
(c) By question a)
E(1/S1 ) = e−µ+σ
2 /2
/S0 .
Note that if we use a random variable to modelize the rate of change from A to B, than
its inverse will be the rate of change from B to A. The product of expected value is always
2
larger than 1 if the variance is non-zero (Jensen’s inequality). Here E(S1 )E(1/S1 ) = eσ .
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Probability and statistics
May 3, 2016
(d) The k-th moment of a lognormal distribution is easy to compute: For k = 1, 2, · · · ,
E(S1k ) = S0k E(ekµ+kσX ) = (S0 eµ )k ek
2 σ 2 /2
.
Q2. Suppose that Z has the standard normal distribution, V has the χ-squared distribution with
n degrees of freedom, and that Z and V are independent. Let
Z
T =p
.
V /n
You will show that T has p.d.f. given by
Γ((n + 1)/2)
f (t) = √
nπΓ(n/2)
−(n+1)/2
t2
1+
n
t ∈ R.
Recall that we have seen in Series 4, that the p.d.f. of a χ-squared with n degrees of freedom
is, for some cn ∈ R:
fV (x) = cn xn/2−1 e−x/2 1{x≥0} .
(a) Find the joint p.d.f. of (T, V ).
(b) Show first that the conditional distribution of T given V = v is normal with mean 0
and variance nv .
(c) Compute cn .
(d) Find the p.d.f. of T .
Solution:
(a) Since Z and V are independent, the p.d.f. of the couple (Z, V ) is
fZ,V (z, v) = fZ (z)fV (v),
√
2
where fZ (z) = e−z /2 / 2π is the p.d.f. of standard normal distribution. By the transformation formula, one obtains the joint p.d.f. of (T, V ):
r
r v
v
fV (v)
fT,V (t, v) = fZ t
.
n
n
(b) The conditional probability given V = v is
r r
p
v/n
fT,V (t, v)
v
v
t2
√
fT |v (t) =
,
= fZ t
= exp −
fV (v)
n
n
2n/v
2π
which is the p.d.f. of a centered normal distribution with variance n/v. One can guess
it by replace directly V by v in the expression of T (which can be proven when Z and
V are independent).
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Probability and statistics
May 3, 2016
(c) The cn can be computed as the constant making fV a p.d.f. (with integral 1):
Z ∞
xn/2−1 exp(−x/2)dx = 2n/2 Γ(n/2).
1/cn =
0
(d) The p.d.f. of T is obtained by integrating fT,V :
Z ∞
fT,V (t, v)dv
fT (t) =
0
Z ∞
√
cn
t2
=√
exp −v
v n/2−1 e−v/2 vdv
2n
2πn 0
2
Z ∞
t
cn
1
exp −v
=√
+
v (n+1)/2−1 dv
2n 2
2πn 0
cn Γ((n + 1)/2)
=√
2πn t2 + 1 (n+1)/2
2n
2
−(n+1)/2
2
Γ((n + 1)/2) −n/2−1/2+(n+1)/2
t
√ 2
+1
=
n
Γ(n/2) πn
2
−(n+1)/2
t
Γ((n + 1)/2)
√ .
=
+1
n
Γ(n/2) πn
Q3. We would like to compute
Z
1
A :=
−3
1
2
√ e−x /2 dx
2π
using Monte-Carlo method:
(a) Express A under the form E(f (U )), where U is a standard Gaussian random variable,
and f an appropriate function.
(b) Take (Ui )i∈N an i.i.d. family having the same law as U . Set
n
Sn :=
1X
f (Ui ).
n i=1
What is the distribution of Sn − A?
(c) Compute E(Sn ) and show that V ar(Sn ) = (A − A2 )/n.
(d) Show that for any x > 0, P(|Sn − A| ≥ x) ≤ 1/nx2 , thus converges to 0 when n → ∞.
(e) Which theorem can you apply to get directly the above convergence?
Solution:
(a) Set f = 1[−3,1] , then
A = E[f (U )] = P(U ∈ [−3, 1]) = P(f (U ) = 1).
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Probability and statistics
May 3, 2016
(b) nSn follows the binomial law with parameter (n, A). Hence
n k
P(nSn = k) =
A (1 − A)n−k ,
k
or equivalently
n k
P(Sn − A = k/n − A) =
A (1 − A)n−k .
k
(c) E(Sn ) = A and
V ar(Sn ) = nV ar(f (Ui )/n)
= (A − A2 )/n.
(d) By Tchebychev inequality
P(|Sn − A| ≥ x) = P(|Sn − A|2 ≥ x2 ) ≤
V ar(Sn )
1
≤
.
2
x
nx2
The convergence is valid for all x > 0, which means that Sn converges in probability
to A. To numerically approximate the value A, one can sample independently a family
(Ui )i=1..n having the standard normal distribution, and counts the number of points
in the interval [−3, 1] then divide by n. While the n becomes larger we get a better
approximation of A.
(e) We can apply the weak Law of large number.
Q4. Fitting a polynomial by Methode of Least Squares Suppose now that instead of simply fitting
a straight line to n plotted points, we wish to fit a polynomial of degree k (k ≥ 2). such a
polynomial will have the following form:
y = β0 + β1 x + β2 x2 + · · · + βk xk .
The method of least squares specifies that the constants β0 , · · · , βk should be chosen that
the sum
n
X
Q(β0 , · · · , βk ) =
[yi − (β0 + β1 xi + · · · + βk xki )]2
i=1
of the squares of the vertical deviations of the points from the curve is a minimum.
(a) Which equation system should a minimizer β̂0 , · · · , β̂k satisfy?
(b) Fit a parabola (polynomial of degree 2) to the 10 points given in the table.
Solution:
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Probability and statistics
May 3, 2016
Table 1: Data for Q4.(b)
i
xi
yi
1
2
3
4
5
6
7
8
9
10
1.9
0.8
1.1
0.1
-0.1
4.4
4.6
1.6
5.5
3.4
0.7
-1.0
-0.2
-1.2
-0.1
3.4
0.0
0.8
3.7
2.0
(a) If we calculate the k + 1 partial derivatives ∂Q/∂β0 , · · · , ∂Q/∂βk , and we set each of
these derivatives equal to 0, we obtain the following k + 1 linear equations involving
k + 1 unknown values β0 , · · · , βk :
n
n
n
X
X
X
k
β̂0 n + β̂1
xi + · · · + β̂k
xi =
yi ,
i=1
β̂0
n
X
xi + β̂1
i=1
n
X
i=1
x2i + · · · + β̂k
n
X
i=1
i=1
xk+1
=
i
i=1
n
X
xi y i ,
i=1
..
.
β̂0
n
X
i=1
xki
+ β̂1
n
X
xk+1
i
+ · · · + β̂k
i=1
n
X
x2k
i
=
i=1
n
X
xki yi .
i=1
As before, if these equations have a unique solution, that solution provides the minimum
value for Q. A necessary and sufficient condition for a unique solution is that the
determinant of the (k + 1) × (k + 1) matrix formed by the coefficients of β̂0 , · · · , β̂k
above is not zero.
(b) In this example, it is found that the equations are
10β0 + 23.3β1 + 90.37β2 = 8.1,
23.3β0 + 90.37β1 + 401.0β2 = 43.59,
90.37β0 + 401.0β1 + 1892.7β2 = 204.55.
The unique solution is
β̂0 = −0.744,
β̂1 = 0.616,
β̂2 = 0.013.
Hence the least squares parabola is
y = −0.744 + 0.616x + 0.013x2 .
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