Expected Wait, Median Wait, and Skew in Geometric Models

Why the Mean and Median Tell Different Stories

When a process repeats until its first success, most waiting times may be short while a few are much longer. A measure of the center should be read with that shape in mind. The mean accounts for every possible wait, including rare long waits; the median marks the point by which at least half of waits have occurred. In a geometric model, these summaries can differ substantially, especially when success on any one trial is unlikely.

We will study the number of trials up to and including the first success. This is a geometric waiting-time model: trials are independent, each trial has the same probability \(p\) of success, and the process continues until a success occurs. Assume \(0<p\leq 1\), and write \(q=1-p\) for the probability of failure on a trial. The random variable \(X\) is the number of trials to the first success, so \(X\) takes the positive integer values \(1,2,3,\ldots\). If \(p=1\), success is certain on the first trial and \(X=1\); the formulas involving logarithms below apply when \(0<p<1\).

Definition: The expected wait is the mean \(E(X)\), the long-run average number of trials to the first success under the model. The lower median wait is the smallest positive integer \(m\) for which \(P(X\leq m)\geq 0.5\). Thus, at least half of all waits are no longer than \(m\), and the integer \(m-1\) does not yet reach that halfway point.

Expected Wait and Median Wait

For a geometric model, the expected wait is \(1/p\). One way to see why is to use the tail probabilities. A positive-integer random variable has expected value equal to the sum of its probabilities of reaching each positive integer threshold: \(E(X)=\sum_{j=1}^{\infty}P(X\geq j)\). Here, reaching trial \(j\) means the first \(j-1\) trials were failures, which has probability \(q^{j-1}\). Summing the geometric series gives the mean.

E(X)=\sum_{j=1}^{\infty}P(X\geq j) =\sum_{j=1}^{\infty}q^{j-1} =1+q+q^2+\cdots =\frac{1}{1-q} =\frac{1}{p}.

The median follows from the cumulative probability. By trial \(m\), the first success has occurred unless the first \(m\) trials were all failures. Therefore \(P(X\leq m)=1-q^m\). Requiring this probability to be at least one-half gives a formula for the lower median. Since \(\ln(q)<0\), dividing by \(\ln(q)\) reverses the inequality.

\begin{aligned} P(X\leq m)\geq 0.5 &\iff 1-q^m\geq 0.5\\ &\iff q^m\leq 0.5\\ &\iff m\ln(q)\leq \ln(0.5)\\ &\iff m\geq \frac{\ln(0.5)}{\ln(q)}. \end{aligned}

Because \(m\) must be a positive integer, when \(0<p<1\) the lower median is the smallest integer at least as large as this logarithmic value:

m=\left\lceil\frac{\ln(0.5)}{\ln(1-p)}\right\rceil.

Worked Example: A Success Probability of 0.20

Suppose a hypothetical process consists of independent trials, each with success probability \(p=0.20\). Let \(X\) be the number of trials through the first success. The geometric model’s assumptions hold by the description: the trials are independent, the probability stays at 0.20, and we count through the first success. Here \(q=1-0.20=0.80\).

The expected wait is \(E(X)=1/p=1/0.20=5\) trials.

The median threshold is \(\ln(0.5)/\ln(0.80)\approx 3.106\), so the smallest integer reaching the threshold is \(m=4\). Check both sides of the cutoff: \(P(X\leq 3)=1-0.80^3=0.488\), which is below 0.5, while \(P(X\leq 4)=1-0.80^4=0.5904\), which is at least 0.5. Thus the median wait is 4 trials, while the expected wait is 5 trials.

Why the Mean Is at Least the Lower Median

For every geometric model with \(0<p\leq 1\), the expected wait is at least the lower median wait. The integer nature of the wait matters: a median must be an integer, while the mean \(1/p\) need not be. Here is a direct argument that does not rely on rounding a rough approximation to the median.

Handle success probabilities of at least one-half.
If \(p\geq 0.5\), then \(P(X=1)=p\geq 0.5\). Thus the lower median is \(m=1\). Also, \(1/p\geq 1\), so \(m\leq E(X)\).

For smaller probabilities, choose an integer threshold.
Suppose \(0<p<0.5\) and let \(n=\lfloor 1/p\rfloor\). Then \(n\geq 2\), \(n\leq 1/p\), and the definition of the floor gives \(p>1/(n+1)\). Therefore \(q=1-p<n/(n+1)\).

Show that the median has been reached by trial \(n\).
We have \(q^n<[n/(n+1)]^n\leq 0.5\). The last inequality follows from \((1+1/n)^n\geq 2\), which is true by the binomial expansion for \(n\geq 2\). Hence \(P(X\leq n)=1-q^n>0.5\), so the lower median satisfies \(m\leq n\).

Compare the integer threshold with the mean.
Since \(n\leq 1/p=E(X)\), we obtain \(m\leq n\leq E(X)\). Together with the first case, this proves the result for all \(0<p\leq 1\).

This comparison does not mean that the mean is a better description of every waiting-time situation. The median describes a halfway point in the distribution; the mean describes its long-run average. When long waits occur, they can pull the mean above the median.

Tail Probabilities Explain the Right Skew

The geometric model’s tail probability is especially useful for understanding its shape. For any nonnegative integer \(k\), \(P(X>k)=q^k\): the wait exceeds \(k\) trials exactly when the first \(k\) trials are failures. The probability at one particular waiting time is \(P(X=k)=q^{k-1}p\). For \(0<p<1\), these individual probabilities decrease as \(k\) increases, while the tail continues to extend to arbitrarily large waits.

This is the characteristic right-skewed pattern: probability is concentrated on smaller waiting times, but there is a tail toward larger values. A small success probability makes \(q\) close to 1, so \(q^k\) decreases slowly. Long waits then retain noticeable probability and can raise the mean relative to the median. A large success probability makes \(q\) small, so the tail shrinks rapidly. The shape remains bounded below at one trial, but long waits become much less likely.

Worked Example: A Small Success Probability

Suppose independent trials have a constant success probability \(p=0.05\), and \(X\) counts trials through the first success. These are the geometric-model assumptions, with \(q=0.95\). The expected wait is \(1/0.05=20\) trials. The median threshold is \(\ln(0.5)/\ln(0.95)\approx 13.513\), so the median is \(m=14\) trials.

The tail probabilities show why the distribution can produce much longer waits than the median. The probability of waiting more than 10 trials is \(P(X>10)=0.95^{10}\approx 0.5987\). The probability of waiting more than 20 trials is \(P(X>20)=0.95^{20}\approx 0.3585\). In context, about 59.87% of waits exceed 10 trials and about 35.85% exceed 20 trials under this model. The mean of 20 trials is not a claim that each wait will be near 20; it is the average across repeated waits, with the extended right tail contributing to that average.

Worked Example: A Large Success Probability

Now suppose independent trials have constant success probability \(p=0.80\), again counting trials through the first success. The model assumptions are the same, but now \(q=0.20\). The expected wait is \(1/0.80=1.25\) trials. The median threshold is \(\ln(0.5)/\ln(0.20)\approx 0.431\), so the lower median is 1 trial. Indeed, \(P(X\leq 1)=0.80\), which already exceeds one-half.

The tail falls quickly: \(P(X>1)=0.20\) and \(P(X>3)=0.20^3=0.008\). Thus only 0.8% of waits exceed three trials under this model. The mean still exceeds the median, but the difference is modest compared with the small-success-probability example. A high probability of success concentrates most of the waiting times near the first trial.

Common Mistake: Treating the Mean as the Typical Wait

A common mistake is to say, “The expected wait is 20 trials, so a typical wait is 20 trials.” That conclusion confuses an average with a halfway point. In the \(p=0.05\) example, the median is 14 trials, and the tail probability \(P(X>20)\) is still about 0.3585. Some waits are much longer, and those large values contribute to the mean. A better interpretation reports both summaries with their meanings: the model has an average wait of 20 trials, and at least half of waits finish by trial 14.

Another error is to infer the median by rounding the mean. The median is determined by the cumulative probability reaching one-half, not by rounding \(1/p\). For \(p=0.20\), for instance, the mean is 5 and the median is 4; for \(p=0.80\), the mean is 1.25 and the median is 1. Use the logarithmic cutoff or check the cumulative probabilities on either side of the proposed median.

Key takeaway: In a geometric waiting-time model, the expected wait is \(1/p\), while the lower median is the smallest integer \(m\) satisfying \(1-(1-p)^m\geq 0.5\). Tail probabilities \(P(X>k)=(1-p)^k\) show how quickly long waits become unlikely. A small success probability leaves a heavier, longer right tail and can make the mean noticeably larger than the median.

Check Your Understanding

Use the geometric-model assumptions and distinguish what the mean, median, and tail probabilities each describe.

For \(p=0.25\), calculate the expected wait and find the lower median using the logarithmic formula. Verify the median by checking the cumulative probability immediately below and at your answer.
In a model with \(p=0.10\), express \(P(X>5)\) in terms of \(q\), then calculate its value to four significant figures.
Explain why a model with a small success probability tends to have a longer right tail than one with a large success probability.
A learner says that a median of 3 trials means every wait is at most 3 trials. Identify the error and state what the median does mean.
In one or two sentences, explain why the geometric mean is at least as large as its lower median.