Solution: 2015 Fall Final - 21

Author: Michiel Smid

Question

Let $n$ be an integer with $n \geq 3$. Consider a bitstring of length $n$, in which each bit is 0 with probability 1/3 (and, thus, 1 with probability 2/3), independently of the other bits. Let $X$ be the number of occurrences of 010 in this bitstring. For example, if the bitstring is $$ 0010100100, $$ then $X = 3$.
What is the expected value $\mathbb{E}(X)$ of $X$?
Hint: Use indicator random variables.

(a)

$2(n-2)/27$

(b)

$n/8$

(c)

$2n/27$

(d)

$(n-2)/8$

Solution

Let $X_i$ be 1 if the bitstring contains 010 and 0 otherwise.

Three is a 1 in 3 chance of a 0 followed by a 2 in 3 chance of a 1 followed by a 1 in 3 chance of a 0.

$ Pr(X_i = 1) = \frac{1}{3} \cdot \frac{2}{3} \cdot \frac{1}{3} $

$ Pr(X_i = 1) = \frac{2}{27} $

When we are predicting the value of the current bit and the next 2 bits, there must be 2 bits after the current bit

As a result, the second last bit and last bit are not considered.

We consider the first $ n-2 $ bits of the bitstring.

$ E(X) = \sum_{i=1}^{n-2} \frac{2}{27} $

$ E(X) = (n-2) \cdot \frac{2}{27} $

$ E(X) = \frac{2(n-2)}{27} $