Contents

% Assignment 1 Solutions

Problem 1.

We use n=12 and obtain the equation $400(1+\beta+\cdots+\beta^{179}) = A(\beta^{180}+\cdots+\beta^{419})$.

format compact     % don't print blank lines between results

b = 1.05^(-1/12);    % (1+rho)^12 = 1 + r_eff
pv_payments = 400*(1-b^180)/(1-b);
% pv_withdraw = A*b^180*(1-b^240)/(1-b)
% find A such that pv_payments = pv_withdraw:
q = b^180*(1-b^240)/(1-b);
A = pv_payments/q

A =
  692.6079

Problem 2(a).

We use n=2 and obtain the equation $500\beta + 600\beta^2 = 1000$.

f = @(b) 500*b + 600*b^2 - 1000;
b = fzero(f,[0,1]);  % Note: we can solve this with formula for quadratic equation
r_eff = b^-2 - 1

r_eff =
    0.1320

Problem 2(b).

We use n=4 and obtain the equation $500\beta^3 + 600\beta^4 = 1000$.

f = @(b) 500*b^3 + 600*b^4 - 1000;
b = fzero(f,[0,1]);
r_eff = b^-4 - 1

r_eff =
    0.1136

Problem 2(c).

We use n=12 and obtain the equation $100(\beta^4+\cdots+\beta^8)+150(\beta^9+\cdots+\beta^{12})=1000$.

f = @(b) 100*b^4*(1-b^5)/(1-b) + 150*b^9*(1-b^4)/(1-b) - 1000;
b = fzero(f,[0,.9999]);
r_eff = b^-12 - 1

r_eff =
    0.1455

Problem 3(a), Method 1

The probability for "heads" is p=0.4, the probability for "tails" is 1-p=0.6.

The random variable Y:=X4 can have the values 0,1,2,3,4. The probabilities are given by

$$P(X_4=k)={4\choose k}p^k(1-p)^{4-k}$$

We then compute $E[Y] = \sum_{k=0}^4 P(Y=k)k$ and $E[Y^2] = \sum_{k=0}^4 P(Y=k)k^2$.

Then $Var[Y] = E[Y^2] - E[Y]^2$.

p = 0.4;
p4 = 1*p^4               % probability for Y=4 (4H, 0T)
p3 = 4*p^3*(1-p)         % probability for Y=3 (3H, 1T)
p2 = 6*p^2*(1-p)^2       % probability for Y=2 (2H, 2T)
p1 = 4*p*(1-p)^3         % probability for Y=1 (1H, 3T)
p0 = (1-p)^4             % probability for Y=0 (0H, 4T)
pv = [p0,p1,p2,p3,p4]    % probabilities for the 5 cases Y=0,...,Y=4
Y = (0:4);               % values of Y in these 5 cases
EY = sum(pv.*Y)          % E[Y]
EY2 = sum(pv.*Y.^2)      % E[Y^2]
VarY = EY2 - EY^2        % Var[Y]

p4 =
    0.0256
p3 =
    0.1536
p2 =
    0.3456
p1 =
    0.3456
p0 =
    0.1296
pv =
    0.1296    0.3456    0.3456    0.1536    0.0256
EY =
    1.6000
EY2 =
    3.5200
VarY =
    0.9600

Problem 3(a), Method 2

We start with the probabilities for X0: P(X0=0)=1. Then we compute the 2 probabilities for X1,..., then the 5 probabilities for X4.

Let Y:=X4, let U:=Y^2. We can then compute E[Y], E[U] by "method 2" explained in the course notes. We first compute the conditional expectations $E[Y\mid X_3=j]$ for all $j=0,\ldots,3$. Then we compute the conditional expectations $E[Y\mid X_2=j]$ for all $j=0,\ldots,2$, etc.

pv = 1;
for k=1:4
  fprintf('k=%g:\n',k)
  pv = 0.6*[pv,0] + 0.4*[0,pv]  % probabilities for X_k=0,X_k=1,...,X_k=k
end

EY = (0:4);               % conditional expectations for Y if X4=0,...,4
EU = (0:4).^2;            % conditional expectations for U if X4=0,...,4

for k=3:-1:0
  fprintf('k=%g:\n',k)
  EY = 0.6*EY(1:end-1) + 0.4*EY(2:end)  % conditional expectations for Y if Xk=0,...,k
  EU = 0.6*EU(1:end-1) + 0.4*EU(2:end)  % conditional expectations for U if Xk=0,...,k
end

VarY = EU - EY^2          % Var[Y]

k=1:
pv =
    0.6000    0.4000
k=2:
pv =
    0.3600    0.4800    0.1600
k=3:
pv =
    0.2160    0.4320    0.2880    0.0640
k=4:
pv =
    0.1296    0.3456    0.3456    0.1536    0.0256
k=3:
EY =
    0.4000    1.4000    2.4000    3.4000
EU =
    0.4000    2.2000    6.0000   11.8000
k=2:
EY =
    0.8000    1.8000    2.8000
EU =
    1.1200    3.7200    8.3200
k=1:
EY =
    1.2000    2.2000
EU =
    2.1600    5.5600
k=0:
EY =
    1.6000
EU =
    3.5200
VarY =
    0.9600

Problem 3(b)

We obtain that $E[Z_1]=p$ and $E[Z_1^2]=p$. Hence

$$Var[Z_1]=E[Z_1^2]-E[Z_1]^2=p-p^2=p(1-p).$$

Since $X_m = Z_1+\cdots+Z_m$ we obtain $E[X_m]=E[Z_1]+\cdots+E[Z_m]=mp$.

Since the random variables $Z_1,\ldots,Z_m$ are independent we obtain

$$Var[X_m] = Var[Z_1]+\cdots+Var[Z_m] = mp(1-p)$$

Note that we obtain for m=4 the same values as above.

p = 0.4;
EZ1 = (1-p)*0 + p*1       % E[Z1] = p
EZ12 = (1-p)*0^2 + p*1^2; % E[Z1^2] = p
VarZ1 = EZ12 - EZ1^2      % Var[Z1] = p - p^2
EY = 4*EZ1
VarY = 4*VarZ1

EZ1 =
    0.4000
VarZ1 =
    0.2400
EY =
    1.6000
VarY =
    0.9600

Problem 3(c)

Find the conditional expectation of X4 under the condition X1=1.

We have 16 possible outcomes: $\omega_1=HHHH,\omega_2HHHT,...,\omega_{16}=TTTT$. We have the probabilities $p_j=P(\{\omega_j\})=p^k (1-p)^{4-k}$ if $\omega_j$ contains k times heads and (4-k) times tails.

The expectation E[Y] is given by $\sum_{\omega_j\in\Omega} p_j Y(\omega_j)$.

For an event $A$ the conditional expectation is given by

$$E[Y\mid A] = \frac{E[1_A\cdot Y]}{P(A)} =\frac{\sum_{\omega_j\in A} p_j\cdot Y(\omega_j)}{\sum_{\omega_j\in A} p_j}$$

Here A is the event $X_1=1$. Hence the first toss must be "heads" and A consists of the 8 outcomes H*** where each "*" is either H or T.

Therefore the possible values for $X_4$ are $1+k$ where $k$ is the number of "heads" in the last three tosses. Hence

$$E[X_4 \mid X_1=1] = 1 + E[X_3] = 1 + 3p = 2.2$$

Method 2: In Problem 3(a), Method 2 we computed actually all conditional expectations $E[Y\mid X_k=j]$ for all $k=3,2,1,0$ and $j=0,\ldots,k$. Therefore we look at the output above: for k=1 we see the two values $E[Y\mid X_1=0] = 1.2$ and $E[Y\mid X_1=1] = 2.2$.


Published with MATLAB® R2016a