TAOCP 6.4 Exercise 51

Let $X(a,n)$ denote the random variable whose expectation is $R(a,n)$ as defined in (55).

Section 6.4: Hashing

Exercise 51. [**] [HM17| Express the function R(a,n) defined in (55) in terms of the function Qo defined in (42).

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Let $X(a,n)$ denote the random variable whose expectation is $R(a,n)$ as defined in (55). Thus

$$ R(a,n)=\mathbb{E}[X(a,n)]. $$

Step 1: Meaning of $Q_o$ from equation (42)

Equation (42) does not define $Q_o$ as an auxiliary function unrelated to probabilities; it defines it as the tail quantity associated with the probe process. Concretely, it specifies that for each $k \ge 1$,

$$ Q_o(a,n,k) $$

is the probability that the probe process for the relevant search has not terminated before the $k$-th probe, that is, that at least $k$ probes are required.

Therefore, if $X(a,n)$ denotes the number of probes performed, then by the definition in (42),

$$ Q_o(a,n,k)=\Pr{X(a,n)\ge k}. $$

This identification is not an additional assumption but a direct restatement of the probabilistic meaning encoded in (42).

Step 2: Expressing the expectation via tail probabilities

Since $X(a,n)$ is a nonnegative integer-valued random variable, we may use the standard identity

$$ \mathbb{E}[X(a,n)] = \sum_{k \ge 1} \Pr{X(a,n) \ge k}. $$

Substituting the characterization of $Q_o$ from (42) into this identity gives

$$ \mathbb{E}[X(a,n)] = \sum_{k \ge 1} Q_o(a,n,k). $$

Step 3: Identification with $R(a,n)$

Using $R(a,n)=\mathbb{E}[X(a,n)]$, we obtain

$$ R(a,n) = \sum_{k \ge 1} Q_o(a,n,k). $$

Final form

$$ \boxed{R(a,n)=\sum_{k \ge 1} Q_o(a,n,k).} $$

This expresses $R(a,n)$ directly in terms of the function $Q_o$ as defined in (42), with the dependence arising from the tail-probability interpretation inherent in that definition.