Problem 5.1

Probabilstic counting

With a $b$-bit counter, we can ordinarily only count up to $2^b - 1$. With R. Morris's probabilistic counting, we can count up to a much larger value at the expense of some loss of precision.

We let a counter value of $i$ represent that a count of $n_i$ for $i = 0, 1, \ldots, 2^b-1$, where the $n_i$ form an increasing sequence of nonnegative values. We assume that the initial value of the counter is 0, representing a count of $n_0 = 0$. The INCREMENT operation works on a counter containing the value $i$ in a probabilistic manner. If $i = 2^b - 1$, then the operation reports an overflow error. Otherwise, the INCREMENT operation increases the counter by 1 with probability $1/(n_{i+1} - n_i)$, and it leaves the counter unchanged with probability $1-1/(n_{i+1} - n_i)$.

If we select $n_i = i$ for all $i \ge 0$, then the counter is an ordinary one. More interesting situations arise if we select, say, $n_i = 2^{i-1}$ for $i > 0$ or $n_i = F_i$ (the $i$th Fibonacci number - see Section 3.2).

For this problem, assume that $n_{2^b-1}$ is large enough that the probability of an overflow error is negligible.

1. Show that the expected value represented by the counter after $n$ INCREMENT operations have been performed is exactly $n$.
2. The analysis of the variance of the count represented by the counter depends on the sequence of the $n_i$. Let us consider a simple case: $n_i = 100i$ for all $i \ge 0$. Estimate the variance in the value represented by the register after $n$ INCREMENT operations have been performed.

Expected value

Suppose at the start of the $j$th increment, the counter holds $i$, which represents $n_i$. If the counter increases, then the value it will increase by $n_{i+1} - n_i$. It happens with probability $1/(n_{i+1} - n_i)$, and so:

\begin{aligned} \E[X_j] &= 0 \cdot \Pr\{\text{stays same}\} + 1 \cdot \Pr\{\text{increases}\} \\ &= 0 \cdot \bigg(1 - \frac{1}{n_{i+1} - n_i}\bigg) + 1 \cdot \bigg((n_{i+1} - n_i) \cdot \frac{1}{n_{i+1} - n_i}\bigg) \\ &= 1 \end{aligned}

This is the expectation any increment. Since there are $n$ increments, the execpted value will be $n$.

Variance

The variance of a single increment.

\begin{aligned} \Var[X_j] &= \E[X_j^2] - \E^2[X_j] \\ &= \bigg(0^2 \cdot \frac{99}{100} + 100^2 \frac{1}{100}\bigg) - 1 \\ &= 99 \end{aligned}

As for the variance of the total value:

$$\Var[X] = \Var[X_1 + X_2 + \ldots + X_n] = \sum_{i=1}^n\Var[X_i] = 99n$$