Question 2 A box contains an unknown number of white and...

Question 2 A box contains an unknown number of white and black balls. We wish to estimate the proportion $p$ of white balls in the box. To do so, we draw $n$ successive balls with replacement. Let $X_n$ be the proportion of white balls obtained after $n$ drawings. (a) What are the mean and variance of $X_n$? (b) Show that, for all $epsilon > 0$, $$P(|X_n - p| ge epsilon) le frac{1}{4nepsilon^2}$$ (c) Using the result in part (b), find a suitable value of $n$ such that, with probability greater than 0.95, the proportion $X_n$ in the sample will estimate $p$ to within an accuracy of 0.1. (d) Now assume that we are in a regime where the Central Limit Theorem applies and we can take $X_n$ to be normally distributed with the mean and variance that you calculated in part (a). Answer the same question as we asked in part (c).

Transcript

00:01 So in this question, we have a box which contains an unknown number of white and black balls.

00:06 So in our box, we have, let's say we have a total of n balls, of which p white n are white, and, well, let's say p n are white, and 1 minus p n are black.

00:34 Now we draw successive balls, so draw n balls with replacement.

00:51 Well, let's say that the number of white balls that we draw, let's say x, is the number of white balls drawn.

01:02 X is going to be binomial with n trials and p chance of success on each trial, because your probability of drawing a white ball is p, because it's p n divided by n.

01:20 So what, actually, let's say that y is this number.

01:29 X n is the proportion of white balls drawn, of proportion of white balls drawn.

01:45 Let's give these y's an index n as well.

01:48 And x n is going to be equal to y n divided by n.

01:53 Now the expected value of y n, we know because it's binomial, is n p, and the variance of y n is n p 1 minus p.

02:03 So that means the expected value of x n is the expected value of y n over n, which is 1 over n times the expected value of y n, which is p.

02:14 And the variance of x n is the variance of y n over n, which is 1 over n squared times the variance of y n, which is p 1 minus p over n.

02:24 So let's just put this in a box.

02:27 Expected value of x n is p, variance of x n is p 1 minus p over n.

02:37 So now let's think about chebyshev's theorem.

02:41 Chebyshev's theorem tells us that if x has an expected value of mu and a variance of sigma squared, then the probability that the modulus of x minus mu is greater than k sigma is less than or equal to 1 over k squared.

03:04 So for x n, the probability that the modulus of x n minus its expected value, which is p, is greater than k times the square root of its variance, which is root p 1 minus p over n, is less than or equal to 1 over k squared.

03:28 So now let's write k root p 1 minus p over n is equal to epsilon.

03:36 Then that means that k is equal to root n over p 1 minus p epsilon.

03:47 So 1 over k squared is equal to p 1 minus p over n epsilon squared.

03:59 But p is between 0 and 1.

04:04 So what is the maximum of this? so let's write d by dp of p 1 minus p is equal to 0.

04:14 This gives us the derivative of p is 1.

04:19 The derivative of minus p squared is minus 2p.

04:23 So this gives us p is equal to a half, at which point a half 1 minus a half is equal to a quarter.

04:33 So that means that p 1 minus p is less than or equal to a quarter on the interval 0 to 1.

04:40 And we know that it's less than or equal to because p 1 minus p evaluated at p equals 0 is 0, and p 1 minus p evaluated at p equals 1 is 0.

04:52 So if p 1 minus p is positive between 0 and 1, then its only stationary point must be a maximum.

05:01 So that means that 1 over k squared is less than or equal to one quarter over n epsilon squared, which is 1 over 4 n epsilon squared.

05:11 So what we have is the probability that the modulus of x n minus p is greater than epsilon, greater than or equal to epsilon, is less than or equal to 1 over 4 n epsilon squared by chebyshev's theorem.

05:34 So now we want to find n such that the probability of estimating a sample within an accuracy of 0 .1, so the difference between x n and p is greater than or equal to 0 .1, is less than 0 .1, so we're within an accuracy of 0 .1, is at least 95 percent.

06:09 Well, that means that the probability that the difference between x n and p is greater than or equal to 0 .1 is going to be, is 1 minus this.

06:20 So if it's 1 minus that, that means that it's less than or equal to 0 .05.

06:32 So what do we have here? well, we have epsilon is equal to 0 .01, and we have 1 over 4 epsilon n squared, sorry, 1 over 4 n epsilon squared is equal to 0 .05, or is less than or equal to 0 .05...

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