Consider the random walk Sn Y1 Y2 Yn where Yis are inde

Consider the random walk Sn = Y1 + Y2 + … + Yn where Yi’s are independently and identically distributed Bernoulli random variables that take value 3 with probability p and the value -2 with probability q = 1 – p.

(i) By trial and error or otherwise, compute the probability P[S5 = 0] if p = 0.6.

(ii) Illustrate the way to compute P[Sk = n], for suitable values of k and n.

(iii) Hence calculate P[S4 = 7] with p = 0.2.

Solution

Sn=Y1+Y2+.....+Yn where Yi\'s are independently and identically distributed Bernoulli random variables that take value 3 with probability p and value -2 with probability q=1-p

as Yi\'s are independently and identically distributed Bernoulli random variables so Sn=Y1+Y2+...+Yn would follow a binomial distribution with paramters n and p

let taking the 3 value may be considered as a success and -2 value as failure

i) now, S5~Bin(5,p) S5=0 means there are two 3 values and three -2 values as 3+3-2-2-2=0

which means there are two successes in 5 trials.

p is given as 0.6 so S5~Bin(5,0.6)

so P[S5=0]=P[there are 2 successes in binomial(5,0.6)]=⁵C₂*0.6²*0.4³=0.2304 [answer]

ii)for Sk=n we have Sk~Bin(k,p)

so to compute P[Sk=n] we have to represent n as a sum of some 3\'s(say x 3\'s) and some -2\'s(say y -2\'s) such that x+y=k and 3x-2y=n.

then P[Sk=n]=P[m successes out of k trials in Bin(k,p)]=^kC_m*p^m*(1-p)^k-m       [answer]

iii) hence we need to calculate P[S4=7] with p=0.2 we have S4~Bin(4,0.2)

we have to represent 7 as a sum of some 3\'s(say x 3\'s) and some -2\'s(say y -2\'s) such that x+y=4 and 3x-2y=7

so we have x=3 and y=1 as 3+3+3-2=7

so P[S4=7]=P[3 succcess in 4 trials in Bin(4,0.2)]=⁴C₃*0.2³*0.8¹=0.0256 [answer]