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166 probability in eecs

function [INVD] = invdist(P)

%This function computes the invariant distribution

% of a stochastic matrix P

N = size(P, 2);

B = (P - eye(N));

B(1:N,N) = ones(N, 1);o = zeros(1, N);

o(N) = 1;

INVD = o*inv(B);

As an example,

P = [0, 0.3, 0.7; 0, 0.4, 0.6; 1, 0, 0];

invdist(P)

produces the invariant distribution

ans =

0.4000 0.2000 0.4000

B.4 Confidence Intervals

We experiment with confidence intervals. Let{Xn, n 1}be i.i.d. with mean and variance 2. We explained in Section 2.3

that a 95% confidence interval for the mean is, provided that n is

sufficiently large,

[n 2n

, n 2n

]

where

n = X1+ +Xn

n

is the sample mean. We test this confidence interval for coin flips.

n +1n

n 1n

n

p

n

n :=

1 + + n

n

95% confidence interval

with n samples

Figure B.3: Confidence intervals forthe bias of a coin using a bound on the

variance.

For coin flips, i.e., B(p)random variables, we know that

var(Xn) = p(1 p) 1/4.

We then use the upper bound 1/2 on the standard deviation in the

confidence intervals that are then given by

[n 1n

, n+ 1

n].

The code below generates i.i.d. B(p)random variables X(1), . . . , X(N)

and computes their sample mean. It then calculates the bounds of the

confidence interval. Figure B.3shows the results.

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\color{blue}

\begin{verbatim}

%Flip coin N times; these are B(p) random variables

%Construct confidence interval

% based on sample mean and bound 1/4 on variance

%N = 300;

p = 0.3;

Y = zeros(1,N);

U = ones(1,N);

M = p*U;

L = zeros(1,N);

Y(1) = (rand(1) < p);

for j = 2:N

Y(j) = ((j - 1)*Y(j-1) + (rand(1) < p))/j;

b = 1/sqrt(j);

U(j) = Y(j) + b;L(j) = Y(j) - b;

end

plot([Y;L;U;M])

Instead of using an upper boundon the confidence interval, one

may use the sample standard deviation.

n:=

X1 + +Xn

n

2

n:=

(X1 n) + + (Xn n)

nn

+ 2nn

n 2

nn

n

Figure B.4: Confidence intervals forthe bias of a coin using the samplevariance.

In that case, the confidence interval is

[n 2nn

, n 2nn

]

where

2n =

X21+ +X2n

n 2n

is the sample variance. Here is the code with the modification. Figure

B.4shows the result.

Initially, the estimate n of the standard deviation is not very good.

However, for large n, this estimate becomes better than an upper

bound and results in smaller confidence intervals.

%Flip coin N times

%Repeat M times

% p = P(1) = 1 - P(0)

N = 300;

p = 0.3;

Y = zeros(1,N); %sample mean

U = ones(1,N); % upper bound of CI

M = p*U; % center of CI

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L = zeros(1,N); % lower bound of CI

S = U; % sample second moment

Y(1) = (rand(1) < p);

S(1) = Y(1)^2;

for j = 2:N

R = (rand(1) < p); % coin flipY(j) = ((j - 1)*Y(j-1) + R )/j;

S(j) = S(j-1) + R^2;

b = 2/sqrt(j);

sigma = sqrt(S(j)/j - Y(j)^2); % samplestandard deviation

U(j) = Y(j) + b*sigma;

L(j) = Y(j) - b*sigma;

end

plot([Y;L;U;M])

The code given above can be used for other distributions. Here is

an example with exponential random variables. Figure B.5shows theresults.

n :=X1 + +Xn

n

2

n:=

(X1 n)2 + + (Xn n)

2

n

n+ 2nn

n 2nn

n

Figure B.5: Confidence intervals for themean of exponential random variablesusing the sample variance.

%Generate N Exp with mean mu

%Constuct confidence interval

% based on sample mean and sample variance

% p = P(1) = 1 - P(0)

N = 300;

mu = 7;

Y = zeros(1,N); %sample mean

U = ones(1,N); % upper bound of CI

M = m u*U; % center of CI

L = zeros(1,N); % lower bound of CI

S = U; % sample second moment

Y(1) = random(exp, mu);

S(1) = Y(1)^2;

for j = 2:N

R = random(exp, mu); % exponential

Y(j) = ((j - 1)*Y(j-1) + R )/j;

S(j) = S(j-1) + R^2;

b = 2/sqrt(j);

sigma = sqrt(S(j)/j - Y(j)^2); % sample standard deviation

U(j) = Y(j) + b*sigma;

L(j) = Y(j) - b*sigma;

end

plot([Y;L;U;M])