In this problem, I want you to do analysis for a biased p, q, with p+q=1. Consid

Question

In this problem, I want you to do analysis for a biased p, q, with p+q=1. Consider a walk of N steps.

p is probability to go to the right and q is probability to go to the left

N+ is Number of steps to go to the right and N- is Number of steps to go to the left

N = N+ + N- , m = N+ - N-

Only right steps (N+ , p) and left steps (N- , q)

<N+> = N*p , <N-> = N*q , <m> = (p-q)*N

In the Binomial distribution, show that <m2> = 4Npq + [(p-q)N]2

3.1. THE PROBABILISTIC FACTS OF LIFE ISTUDENT VERSION, DECEMBER 8, 2002] 69 farther away from ro before the factore-o) /20 begins to hurt. Remembering that P(x) is a probability distribution, this observation means that for bigger you're likely to find measurements with bigger deviations from the most-likely value ro. The prefactor of 1/a in front of Equation 3.8 arises because a wider bump (larger ) needs to be lower to maintain a fixed area. Let's make all these ideas more precise, for any kind of distribution m xo before the factor e 3.1.3 Mean and variance The average (or mean or expectation value) of r for any distribution is written (x) and defined discrete (3.9) dx xP(x) , continuous For the uniform and Gaussian distributions, the mean is just the center point. That's because these distributions are symmetrical: There are exactly as many observations a distance d to the right of the center as there are a distance d to the left of center. For a more complicated distribution this needn't be true; moreover, the mean may not be the same as the most probable value, which is the place where P(x) is maximum. More generally, even if we know the distribution of x we may instead want the mean value of some other quantity f(x) depending on x. We can find via ,f(xi)P(x) Jdx f(x)P() , discrete continuous (3.10) If you go out and measure x just once you won't necessarily get 2) right on the nose. There is some spread, which we measure using the root-mean-square deviation (or RMS deviation, or standard deviation) deviation 2 Example f f a. Show thatf) b. Show that if the RMS deviation equals zero, this implies that every measurement of x really does give exactly 2) Solution a. We just note that f) is a constant (that is, a number), independent of x. The average of a constant is just that constant b. In the formula 0 = (x-(z))2- P(xi)(zi-(z))2, the right h have any negative terms. The only way this sum could equal zero is for every ternm to be zero separately, which in turn requires that P(a) = 0 unless xi-2) or any function of X and side doesn T Note that it's crucial to square the quantity (x - (x)) when defining the RMS deviation; otherwise we'd trivially get zero for the average value (x - (x))). Then we take the square root just to get something with the same dimensions as x. We'll refer to((-(x))2) as the variance of x (some authors call it the second moment of P(x)) Your Turn 3b a. Show that variance(z) = x2)-((2))2 b. Show for the uniform distribution (Equation 3.5) that variance(/12

Explanation / Answer

here <x> means expectation of X that is E[X]

and <<x-<x>>2> means variance of X that is V(X)

N+ denote the number of right steps.. so N+ ~Bin(N,p)

N- denotes the number of left steps   N-~ Bin(N,q)

now N=N+ + N-

m=N+ - N-

E[N+]=pN

E[N-]=qN

so E[m]=E[N+]-E[N-]=(p-q)N

we need to find E[m2]

now we know from standard formula V(m)=E[m2]-E[m]2

so E[m2]= V(m)+E[m]2

now V[m]=V[N+ - N-]=V[N+ - (N-N+)]   [as N=N+ + N-]

                               =V[2N+ -N]=V[2N+] as N is a constant and variance does not depend upon change of origin.

                               =4V[N+]=4pqN   [as N+~Bin(N,p) so V[N+]=Npq where q=1-p]

and we know E[m]=(p-q)N so E[m]2=[(p-q)N]2

so E[m2]= V(m)+E[m]2=4pqN+[(p-q)N]2

or <m2>=4Npq+[(p-q)N]2    [proved]

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