Question

Please Solve this question by using Matlab.

errors accumulate ran 1.1.6. It's reasonable to expect that floating point during a long computation, creating what is known as a random walk. O average we expect as many errors to be negative as positive, so they tend to partially cancel out. Suppose we define a random sequence by x 0 and 4-1 ± 1 for n 1, with the signs chosen by tossing a fair coin for each n. Let a, and B, be the average values of x, and I.,), respectively, over all such waks Then a classic result of probability is that a, 0 and (a) In MATLAB the function randn simulates drawing numbers from the normal or Gaussian distribution G.e., the bell curve) with mean zero and variance 1. Choose a unique positive integer seed value s (for example, use the last 5 digits of your phone number) and enter rng(s) to initialize the random number generator. Then the following code generates one random walk for n 10: r randn (10000,1); % draw random numbers % cumulative summation cunsum(sign(r)); We can plot every fifth number in the sequence using plot(x(1:5:end)), hold on Plot 50 such random walks all together on one graph. x (b) Perform a million random walks, computing the average values oft oe /2n/r at n-: 10000. and mooool. Compare these to α" ~O and ß"

Answer 1

clear
%a
for i=1:50
r=randn(10000,1);
x=cumsum(sign(r));
plot(x(1:5:end)),hold on
end
%b
xn=zeros(1,1000000);
xn_abs=xn;
for i=1:1000000
r=randn(10000,1);
x=cumsum(sign(r));
xn(i)=x(end);
xn_abs(i)=abs(x(end));
end
fprintf('average of x(10000): %f\n',mean(xn))
fprintf('average of |x(10000)|: %f\n',mean(xn_abs))
fprintf('Bn: %f\n',sqrt(2*10000/pi))
fprintf('an=0\n')

$Command Window MAT EDITOR PUBLISH > randomwa average of x(10000): 0.014472 average of x (10000) 79.693844 Bn: 79.788456 an-0>> Find Files New Open Save Compare ▼ Go To▼ D Breakpoints Run Run and Print Find Advance ▼ FILE NAVIGATE Figure 1 1clear File Edit View Tools Desktop Window Help 3- El for i=1 : 50 r=randn (10000,1) ; x-cumsum (sign (r)); plot (x (1:5:end)) , hold on 250 end 200 9-xn-zeros (1, 1000000) abs=xn ; 10- xn 150 11-for i-1:1000000 r=randn (10000,1) ; x-cumsum (sign (r)) xn (i)-x (end); xn abs (i)-abs (x(end) ); 100 13- 14 15- 16 - 17- fprintf ( average of x(10000): %f\n,mean (xn)) 18- fprintf ( averageoflx ( 10000)I:%f\n,mean/xn 19- tprintf (Bn: %f\n, sart (2*10000/pi)) 20fprintf (an-0 n) 50 end abs )) -50 100 -150 200 0 200 400 600 800 1000 1200 1400 1600 1800 2000$