67) E- At all times, an urn contains N balls --- some white ballsand some black

Question

67) E- At all times, an urn contains N balls --- some white ballsand some black balls. At each state, a coin havingprobability p, 0< p < 1, of landing heads is flipped. If heads appears, then a ball is chosen at random from the urn andis replaced by a white ball; if tails appears, then a ball ischosen from the urn and is replaced by a black ball. LetXn denote the number of white balls in the urn after thenth stage.

If p=1, what is the expected time until there are only white ballsin the urn if initially there are i white and N-i black?

Explanation / Answer

This is a chain.

Since p = 1 each time we replace the ball only with whiteball.

Let us write the expected number of white ball in aftern+1 draws as a function of Xn and N.

Probability of drawing a while ball from Xn while ball and totalN balls is Xn/N. When we draw a while ball number ofwhite balls remains same when we draw a black ball number of whiteballs will increase by 1.

So E(Xn+1) = Xn *(Xn/N)+(Xn + 1)(1- Xn/N )

E(X2 ) = 1- i/N

E(X3) = 1- (1-i/N)/N = 1-1/N+i/N2

E(X4) = 1- (1-1/N+ i/N3)/N

Therefore E(Xn) =1-1/N+1/N2-1/N3 + ----+/-(1-)n+1*i/Nn-1

We need to write this as a sum of three terms to be able tosolve this.

First we know that 1-1/N+1/N2- ….. sum ofinfinite series = 1/(1-1/N) = N/(N-1) .

We write E(Xn) =1-1/N+1/N2-1/N3 +----+(-1)n-2*1/Nn-2+(1-)n-1*i/Nn-1

= (1-1/N+1/N2-1/N3 +----+(-1)n-2*1/Nn-2……….Sum of infinite series) –((-1)n-2*1/Nn-2+………..sum of infinite series)+(1-)n-1*i/Nn-1

= N/(N-1) + ((-1)n-2*1/Nn-2 )*(N/N-1) + (1-)n-1*i/Nn-1

Now that E(Xn) = 0

ð N/(N-1) + ((-1)n-2*1/Nn-2 ) *(N/N-1) +(1-)n-1*i/Nn-1 = 0

Let us solve it in two cases case 1: n is odd

ð N/(N-1) –N3-n/(N-1)+i/ Nn-1 = 0

ð N/(N-1) +i/ Nn-1 = N3-n/(N-1)

This has no solution for N becausel.h.s >1(because first term >1 and second term is positiveterm) and R.H.S is <1 for all N>2. And not defined for N=1 .So n must be even.

Now if n is even

ð N/(N-1) + ((-1)n-2*1/Nn-2 ) *(N/N-1) +(1-)n-1*i/Nn-1 = 0

ð N/(N-1)+N3-n/(N-1)= i/ Nn-1

ð Nn+N2=i(N-1)

ð Nn = i(N-1) - N2

Taking ln on both sides

ð ,nln(N) = ln(i(N-1) - N2 )

So n = ln(i(N-1) - N2 )ln(N)

Hope this helps you. Feel free to ask for anyclarifications.

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