The negative binomial, lognormal and Pareto distributions, derived

X = x means the r-th success arrives on trial x exactly. So the last trial is a success, and the first x − 1 trials contain exactly r − 1 successes, in any order. A worker applies for jobs and each application succeeds with chance 0.25. How many applications until the third offer? Take x = 5: the fifth application is an offer and the first four contain two offers.

The r − 1 earlier successes can fall among the first x − 1 trials in C(x − 1, r − 1) ways. Every such sequence has r successes and x − r failures, so each has probability p^rq^x−r. The sequences are exclusive, so the probabilities add:

For r = 3, p = 0.25 and x = 5: C(4, 2) × 0.25³ × 0.75² = 6 × 0.015625 × 0.5625 = 0.0527. The probabilities over x = 3, 4, 5, … add to 1. With r = 1 this is the geometric distribution, the wait for the first success.

The wait for the r-th success is the wait for the first, plus the further wait for the second, and so on: r independent geometric waits, each with mean 1 ÷ p and variance q ÷ p². Means add and, for independent waits, variances add:

For r = 3 and p = 0.25 that is 12 applications on average, with variance 36. The same product idea gives the moment generating function, the r-th power of the geometric one: ${\left[\frac{p{e}^t}{1-q{e}^t}\right]}^r$, as in the MGF derivation.

The r-th success has arrived by trial x exactly when the first x trials contain at least r successes, so P(X ≤ x) = P(Binomial(x, p) ≥ r). The lab shows the two numbers agreeing. Some books count instead the failures before the r-th success, Y = X − r, with mean rq ÷ p (here 9) and the same variance. Check which version a question uses before substituting.

Incomes change by percentages: a raise of 5% multiplies income by 1.05. A product of many such factors becomes a sum after taking logs, and a sum of many small independent pieces is close to normal by the central limit theorem. So when growth compounds, ln Y is normal and Y is lognormal. The sixty Ashanagar incomes behave this way: the mean of ln (income in thousands) is 3.253 and its standard deviation is 0.471.

Y ≤ y exactly when X ≤ ln y, so the distribution function is P(Y ≤ y) = Φ((ln y − μ) ÷ σ). The density is its slope. The slope of ln y is 1 ÷ y, which brings in the extra factor:

The curve starts at zero, rises to a peak and has a long tail to the right, since y can be any positive number.

E(Y^k) = E(e^kX), which is the moment generating function of the normal at t = k, namely e^{kμ + k²σ²/2} (MGF derivation). Put k = 1 and k = 2:

For the Ashanagar fit the mean is ₹28.9 thousand, close to the households’ actual 28.8.

The median: P(Y ≤ e^μ) = P(X ≤ μ) = ½, so the median is e^μ. The mode: set the slope of ln f(y) = −ln y − (ln y − μ)² ÷ (2σ²) to zero, which gives −1 ÷ y − (ln y − μ) ÷ (σ²y) = 0, so ln y = μ − σ² and the mode is e^μ−σ². Since σ² > 0, e^μ−σ² < e^μ < e^μ+σ²/2. For Ashanagar: mode 20.7, median 25.9, mean 28.9. A few high incomes pull the mean up, and they do so more as σ grows: the mean is e^σ²/2 times the median.

The Lorenz curve of a lognormal is L(p) = Φ(Φ⁻¹(p) − σ), and integrating it gives a Gini coefficient that depends on σ alone, G = 2Φ(σ ÷ √2) − 1. The scale μ does not matter: doubling every income leaves inequality unchanged. For σ = 0.47 it gives 0.261, against 0.256 computed straight from the sixty incomes. The top tenth of households hold 20.9% by the model. We state the integral and check it against simulated data. The Gini guide builds the Lorenz curve from scratch.

Pareto noticed that the number of people with income above x falls by a constant factor each time x is doubled, whatever x is. With P(X > x) = (x_m ÷ x)^α, doubling x multiplies the survival chance by 2^−α. That is scale-freedom: nothing in the shape depends on the unit of rupees. x_m is the smallest income and α, the tail index, says how fast the tail falls. On log scales P(X > x) is a straight line of slope −α. The density is the negative slope of the survival function:

Integrate x times the density:

For α ≤ 1 the integral of x^−α diverges, so there is no mean at all. The same calculation for E(X²) needs α > 2, giving variance αx_m² ÷ ((α − 1)²(α − 2)). With x_m = 10 and α = 2 the mean is 20, the median is 10 × 2^1/2 = 14.1, and the variance is infinite: averages of such incomes converge to the mean very slowly, if at all.

The fraction of people above x is q = (x_m ÷ x)^α. Their total income is the integral of t f(t) from x, which is αx_m^αx^1−α ÷ (α − 1). Divide by the mean αx_m ÷ (α − 1):

The 80/20 rule, the top 20% holding 80%, asks 0.2^1−1/α = 0.8, which gives α = ln 5 ÷ ln 4 = 1.161. With α = 2 the top 10% hold 0.1^0.5 = 31.6%.

The shares above give the Lorenz curve directly. The poorest fraction p = 1 − q holds 1 − q^1−1/α, so L(p) = 1 − (1 − p)^1−1/α. The Gini coefficient is 1 − 2∫L(p)dp:

A thinner tail (larger α) means less inequality: α = 2 gives G = 0.333, α = 3 gives 0.2. The lab draws the density and the straight log-log line and reads off these shares.