Suppose that $X_1,X_2,...$ are real numbers between 0 and 1 that are chosen independently and uniformly at random. Let $S=\sum_{i=1}^k X_i/2^i$, where $k$ is the least positive integer such that $X_k<X_{k+1}$, or $k=\infty$ if there is no such integer. Find the expected value of S.
user
Sure, I'd be happy to help with this problem.

This is a classic example of a random sum problem known as the distribution of the sum of the first k positive rational numbers. The expected value of S, denoted as E[S], is given by:

E[S] = 0.5(1 - 2^(-k))

This formula can be derived by considering all possible cases for k:

1. If k = 0, then S = 0.5, so E[S] = 0.5.

2. If 1 ≤ k < ∞, then S = 0.5 + 2^(-1)X_1 + 2^(-2)X_2 + ... + 2^(-k)X_k.

Take the expected value of this expression:

E[S] = 0.5 + 2^(-1)E[X_1] + 2^(-2)E[X_2] + ... + 2^(-k)E[X_k]

= 0.5 + (1/2) + (1/4) + ... + (1/2^k)

= 1 - 2^(-k)

Multiplying both sides by 2^k, we get:

2^kE[S] = 2^k - 1

Dividing both sides by 2^k:

E[S] = 0.5(1 - 2^(-k))

3. If k = ∞, then S = 0.5 + 2^(-1)X_1 + 2^(-2)X_2 + ..., so the expected value is undefined.

This result means that if there is no smallest positive rational number in the sequence, then the expected value of S is undetermined.

In your specific scenario, if X_1, X_2, ... Are independently and uniformly distributed between 0 and 1, then for any positive integer k, E[X_{k}] = (1/2) and E[X_{k+1}] = (1/2) + (1/2^(k+1)) = 1/2 * (1 + (1/2^k)), so:

E[S] = 0.5 [1/2 * (1 + (1/2^k)) - 2^(-k)] = 0.5 - 0.5 * 2^(-k)

This is the same formula that we obtained in case 2.

Therefore, the expected value of S is:

E[S] = 0.5(1 - 2^(-k))

I hope this helps!
assistant
Suppose that $X_1,X_2,...$ are real numbers between 0 and 1 that are chosen independently and uniformly at random. Let $S=\sum_{i=1}^k X_i/2^i$, where $k$ is the least positive integer such that $X_k<X_{k+1}$, or $k=\infty$ if there is no such integer. Find the expected value of S.
user
This is a complex problem that involves understanding of stochastic processes and expectations. Let's break it down.

First, let's define $k$ as a random variable that represents the least positive integer such that $X_k < X_{k+1}$. The probability that $k=n$ is the probability that the first $n$ numbers are decreasing and the $(n+1)^{th}$ number is larger than the $n^{th}$ number. Since the numbers are chosen uniformly and independently, this probability is $1/(n+1) * 1/2$.

Now, let's consider the expected value of $S$ conditioned on $k=n$. This is the expected value of the sum of the first $n$ numbers divided by $2^i$. Since each $X_i$ is uniformly distributed between 0 and 1, the expected value of $X_i/2^i$ is $1/2^{i+1}$. Therefore, the expected value of $S$ conditioned on $k=n$ is $\sum_{i=1}^{n} 1/2^{i+1} = 1/2 - 1/2^{n+1}$.

Now, the expected value of $S$ is the sum over all possible values of $k$ of the expected value of $S$ conditioned on $k=n$ times the probability that $k=n$. This gives us $E[S] = \sum_{n=1}^{\infty} (1/2 - 1/2^{n+1}) * 1/(n+1) * 1/2$.

This is a complex sum that doesn't have a simple closed form. However, it can be approximated numerically.
assistant