Question

In Section 3.6.3, we saw that if \(U\) is a random variable that is uniform on \((0,1)\) and if, conditional on \(U=p, X\) is binomial with parameters \(n\) and \(p\), then $$ P\\{X=i\\}=\frac{1}{n+1}, \quad i=0,1, \ldots, n $$ For another way of showing this result, let \(U, X_{1}, X_{2}, \ldots, X_{n}\) be independent uniform \((0,1)\) random variables. Define \(X\) by \(X=\\# i: X_{i}

Short answer

In summary, we found that the probability of X having exactly i values smaller than U is given by \( P\{X=i\}=\binom{n}{i}U^i(1-U)^{n-i} \), and the expected value of this probability is \( E[P\{X=i\}]=\frac{1}{n+1} \). This result is the same as the one provided in Section 3.6.3, which proves the relationship between this problem and the conditional binomial distribution based on a uniform random variable U.

Step-by-step solution

Finding the probability of X having a particular value

Since X is defined as the number of Xi values smaller than U, we want to calculate the probability of exactly i Xi values being smaller than U while the rest are larger than U. This event can be represented as follows: \(P\{X = i\} = P\{X_1 < U, X_2 < U, \ldots, X_i < U, X_{i+1} > U, \ldots, X_n > U\}\) We can first calculate the individual probabilities of each condition given that these random variables are uniform on (0,1): \( P\{X_j < U\} = P\{U - X_j > 0\} = U\) \(P\{X_k > U\} = P\{X_k - U > 0\} = 1 - U\)

Calculating the probability for all cases

Using the fact that all the random variables are independent, we can calculate the probability of all these conditions happening simultaneously by multiplying the probabilities of each condition: \(P\{X = i\} = \binom{n}{i} U^i (1 - U)^{n - i}\), where \(\binom{n}{i}\) is the number of possible combinations of choosing i elements from an n-element set. Now, we should find the expected value of this probability: \(E[P\{X = i\}] = \int_0^1 \binom{n}{i} U^i (1 - U)^{n - i} dU\)

Evaluate the integral

We will now evaluate the integral with respect to U: \(E[P\{X = i\}] = \binom{n}{i} \int_0^1 U^i (1 - U)^{n-i} dU\) This integral is equivalent to calculating the beta function: \(B(a, b) = \int_0^1 x^{a-1}(1-x)^{b-1}dx\), where a = i+1 and b = n-i+1. Using properties of the beta function, we know that \(B(a, b) = \frac{(a-1)!(b-1)!}{(a+b-1)!}\)

Calculate the expected value

Substitute the beta function into the expected value expression: \(E[P\{X = i\}] = \binom{n}{i} \frac{i!(n-i)!}{(n+1)!}\) Notice that this simplifies to: \(E[P\{X = i\}] = \frac{1}{n+1}\)

Relate to given result and conclude

Now that we have found \(E[P\{X = i\}] = \frac{1}{n+1}\), we see that it is equal to the provided result. This is consistent with the conditional binomial distribution given in the problem, as the probability \(P\{X = i\}\) also accounts for all possible values of p from the uniform random variable U.