Chapter 8 Frequently Used Continuous Distributions

8.1 Choosing a continuous model

Continuous families describe measurements such as time, loss severity, return, and proportion. Their support, symmetry, tail behavior, and generating mechanism matter more than superficial visual resemblance.

Each explorer plots the density and cdf side by side and allows a second parameter set to be superposed. Watch how location, spread, skewness, and cumulative probabilities change together.

8.2 Continuous uniform distribution

Why it is useful—two applications. (1) Model an arrival time known only to fall somewhere in a fixed interval when every subinterval of equal length is equally plausible. (2) Represent rounding error over \((-0.5,0.5)\) when values are rounded to the nearest integer. The bounded, flat density expresses absence of preference within known limits.

\(X\sim\operatorname{Uniform}(a,b)\) has density

\[f_X(x)=\frac{1}{b-a},\quad a<x<b,\]

with \(\mathbb{E}(X)=(a+b)/2\) and \(\operatorname{Var}(X)=(b-a)^2/12\).

The model assigns equal probability to intervals of equal length. It is useful for bounded uncertainty when no location inside the interval is preferred.

8.3 Normal distribution

Why it is useful—two applications. (1) Model measurement error created by many small independent disturbances. (2) Describe biological measurements such as adult height within a sufficiently homogeneous population. Its central role comes from both its symmetric shape and the tendency of standardized sums and averages to become approximately normal.

\(X\sim N(\mu,\sigma^2)\) has density

\[f_X(x)=\frac{1}{\sigma\sqrt{2\pi}} \exp\!\left[-\frac{(x-\mu)^2}{2\sigma^2}\right].\]

\(\mu\) moves the center and \(\sigma\) controls horizontal spread. The distribution is symmetric, so mean, median, and mode equal \(\mu\). About 68%, 95%, and 99.7% of probability lies within one, two, and three standard deviations.

Standardization

If \(X\sim N(\mu,\sigma^2)\), then

\[Z=\frac{X-\mu}{\sigma}\sim N(0,1).\]

Standardization expresses a value in standard-deviation units and lets every normal probability be calculated from one reference distribution.

De Moivre–Laplace theorem

The normal approximation to a binomial count is not only a convenient numerical rule. It follows from a classical limit theorem.

Let \(X_n\sim\operatorname{Binomial}(n,p)\) with fixed \(0<p<1\). As \(n\to\infty\),

\[\frac{X_n-np}{\sqrt{np(1-p)}}\xrightarrow{d}N(0,1).\]

Equivalently, for real numbers \(a<b\),

\[\mathbb{P}\!\left(a\le \frac{X_n-np}{\sqrt{np(1-p)}}\le b\right) \longrightarrow \Phi(b)-\Phi(a).\]

The theorem says that after centering a binomial count at its mean \(np\) and measuring distance in standard deviations \(\sqrt{np(1-p)}\), its distribution becomes approximately standard normal. The approximation improves as both \(np\) and \(n(1-p)\) become large.

Discrete probabilities occupy bars of width 1, whereas a normal distribution assigns probability over continuous intervals. Therefore, use a continuity correction when applying the theorem. For example, \(\mathbb{P}(X\le k)\) is approximated by \(\mathbb{P}(Y\le k+0.5)\), not \(\mathbb{P}(Y\le k)\). Similarly, \(\mathbb{P}(a\le X\le b)\) is approximated using the interval from \(a-0.5\) to \(b+0.5\).

The approximation is especially useful for large binomial counts, such as the number of voters supporting an option in a large sample, because it replaces a lengthy discrete sum with a standardized normal probability.

Example: public-opinion responses

Suppose 1,000 independently sampled respondents support a proposal with probability \(p=0.55\). If \(X\) is the number supporting it, then

\[\mu=np=550,\qquad \sigma=\sqrt{np(1-p)}\approx15.73.\]

Using a continuity correction,

\[\mathbb{P}(530\le X\le570) \approx \Phi\!\left(\frac{570.5-550}{15.73}\right) -\Phi\!\left(\frac{529.5-550}{15.73}\right).\]

8.4 Log-normal distribution

Why it is useful—two applications. (1) Model repair costs, household incomes, file sizes, and environmental concentrations when values are positive and strongly right-skewed. (2) Model quantities produced by multiplicative growth: if many small percentage changes multiply together, their logarithms add and can be approximately normal. The log-normal is therefore useful when unusually large values are possible but negative values are impossible.

A positive random variable \(X\) is log-normal with parameters \(\mu\) and \(\sigma^2\) when

\[\log X\sim N(\mu,\sigma^2).\]

Equivalently,

\[f_X(x)=\frac{1}{x\sigma\sqrt{2\pi}} \exp\!\left[-\frac{(\log x-\mu)^2}{2\sigma^2}\right], \quad x>0.\]

Its principal summaries are

\[\mathbb{E}(X)=e^{\mu+\sigma^2/2},\qquad \operatorname{Var}(X)=(e^{\sigma^2}-1)e^{2\mu+\sigma^2},\]

\[\operatorname{Median}(X)=e^\mu, \qquad \operatorname{Mode}(X)=e^{\mu-\sigma^2}.\]

The parameters describe the distribution on the logarithmic scale, not the original scale. Increasing \(\mu\) shifts the distribution toward larger positive values. Increasing \(\sigma\) produces more right-skewness and a longer upper tail. Unlike the normal distribution, the mean exceeds the median when \(\sigma>0\).

Real-life example: equipment repair time

Suppose repair time \(X\) in minutes is log-normal with \(\mu=\log(30)\) and \(\sigma=0.5\). The median repair takes 30 minutes, but the mean is

\[\mathbb{E}(X)=30e^{0.5^2/2}\approx33.99\text{ minutes}.\]

The probability that a repair lasts more than one hour is

\[\mathbb{P}(X>60) =1-\Phi\!\left(\frac{\log(60)-\log(30)}{0.5}\right) \approx0.083.\]

The gap between the median and mean reflects occasional long repairs. This is practically important for staffing: the median describes a typical job, while the mean determines total expected workload.

8.5 Exponential distribution

Why it is useful—two applications. (1) Model the waiting time to the next call in a constant-rate call stream. (2) Model the lifetime of an electronic component during a period in which its instantaneous failure rate is approximately constant. Its memorylessness is both its attraction and its main limitation.

\(X\sim\operatorname{Exponential}(\lambda)\) has

\[f_X(x)=\lambda e^{-\lambda x},\quad x>0,\]

\[F_X(x)=1-e^{-\lambda x},\qquad S_X(x)=e^{-\lambda x}.\]

Its mean is \(1/\lambda\) and variance is \(1/\lambda^2\).

The exponential models time to an event under a constant hazard \(h(x)=\lambda\).

Memoryless property

If \(X\sim\operatorname{Exponential}(\lambda)\), then for all \(s,t\ge0\),

\[\mathbb{P}(X>s+t\mid X>s)=\mathbb{P}(X>t).\]

Indeed,

\[\mathbb{P}(X>s+t\mid X>s) =\frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} =e^{-\lambda t}.\]

The interpretation is subtle: after the event has not occurred for \(s\) time units, the remaining waiting time has the same exponential distribution as the original waiting time. The system has no statistical “age.” This is realistic only when the event rate remains constant; it is unsuitable for equipment whose failure risk increases through wear.

Real-life example: waiting for the next help-desk call

Suppose calls arrive at a help desk according to a constant-rate Poisson process with mean one call every 10 minutes. The time \(X\) to the next call is exponential with rate \(\lambda=0.1\) per minute. An employee has already waited 8 minutes without receiving a call. The probability of waiting at least 5 additional minutes is

\[\mathbb{P}(X>13\mid X>8)=\mathbb{P}(X>5)=e^{-0.1(5)}\approx0.6065.\]

The elapsed eight minutes do not make a call more “due”; under the constant-rate model, the future waiting time restarts probabilistically.

8.6 Gamma distribution

Why it is useful—two applications. (1) Model the waiting time until the \(\alpha\)th event in a Poisson process, such as the time until the fifth arrival. (2) Model positive, right-skewed quantities such as rainfall accumulation or task-completion time. The shape parameter lets the density range from strongly skewed to nearly symmetric.

Using shape \(\alpha\) and rate \(\lambda\), \(X\sim\operatorname{Gamma}(\alpha,\lambda)\) has

\[f_X(x)=\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x},\quad x>0,\]

with \(\mathbb{E}(X)=\alpha/\lambda\) and \(\operatorname{Var}(X)=\alpha/\lambda^2\).

When \(\alpha\) is an integer, a gamma variable can represent the waiting time to the \(\alpha\)th event of a Poisson process. It is also flexible for positive, right-skewed insurance losses.

Software may use a scale \(\theta=1/\lambda\) instead of a rate. Always check the convention: under the scale convention, the mean is \(\alpha\theta\).

8.7 Beta distribution

Why it is useful—two applications. (1) Model an unknown probability, such as the success rate of a new teaching method, because the support is exactly \((0,1)\). (2) Model a random proportion such as the fraction of a habitat covered by vegetation. Its two shape parameters can express symmetry, skewness, concentration, or U-shaped behavior.

\(X\sim\operatorname{Beta}(\alpha,\beta)\) has support \(0<x<1\) and density

\[f_X(x)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}.\]

Its mean is \(\alpha/(\alpha+\beta)\) and variance is

\[\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}.\]

The beta family is natural for proportions and probabilities. Its two shape parameters produce uniform, symmetric, skewed, U-shaped, or highly concentrated densities.

8.8 Summary of continuous distributions

Distribution	Support	Typical use	Mean	Variance
Uniform\((a,b)\)	\((a,b)\)	Bounded equal-density uncertainty	\((a+b)/2\)	\((b-a)^2/12\)
Normal\((\mu,\sigma^2)\)	\(\mathbb{R}\)	Symmetric measurements and sums	\(\mu\)	\(\sigma^2\)
Log-normal\((\mu,\sigma^2)\)	\((0,\infty)\)	Positive multiplicative, right-skewed quantities	\(e^{\mu+\sigma^2/2}\)	\((e^{\sigma^2}-1)e^{2\mu+\sigma^2}\)
Exponential\((\lambda)\)	\((0,\infty)\)	Time to first constant-rate event	\(1/\lambda\)	\(1/\lambda^2\)
Gamma\((\alpha,\lambda)\)	\((0,\infty)\)	Waiting time or positive severity	\(\alpha/\lambda\)	\(\alpha/\lambda^2\)
Beta\((\alpha,\beta)\)	\((0,1)\)	Proportions and probabilities	\(\alpha/(\alpha+\beta)\)	\(\alpha\beta/[(\alpha+\beta)^2(\alpha+\beta+1)]\)

Exercises

Exercise 1: uniform

\(X\sim\operatorname{Uniform}(100,500)\). Find \(\mathbb{P}(X>350)\) and the 90th percentile.

Show solution

\(\mathbb{P}(X>350)=(500-350)/400=0.375\). The 90th percentile is \(100+0.9(400)=460\).

Exercise 2: normal

A claim-processing time is normal with mean 30 minutes and standard deviation 5 minutes. Find \(\mathbb{P}(25<X<40)\).

Show solution

\[\Phi(2)-\Phi(-1)\approx0.8186.\]

Exercise 3: normal approximation

Approximate \(\mathbb{P}(X\le55)\) for \(X\sim\operatorname{Binomial}(100,0.5)\).

Show solution

Use \(Y\sim N(50,25)\) and continuity correction:

\[\mathbb{P}(X\le55)\approx\Phi\!\left(\frac{55.5-50}{5}\right)=\Phi(1.1).\]

Exercise 4: exponential

A lifetime is exponential with mean 10 years. Find \(\mathbb{P}(X>15\mid X>5)\).

Show solution

By memorylessness this is \(\mathbb{P}(X>10)=e^{-1}\).

Exercise 5: gamma

If \(X\sim\operatorname{Gamma}(3,0.5)\) using the rate convention, find its mean, variance, and \(\mathbb{P}(X>8)\).

Show solution

The mean is 6, the variance is 12, and

\[\mathbb{P}(X>8)=1-F_X(8)\approx0.2381.\]

Exercise 6: beta

If \(X\sim\operatorname{Beta}(2,5)\), find its mean and \(\mathbb{P}(X<0.4)\).

Show solution

The mean is \(2/7\) and

\[\mathbb{P}(X<0.4)=F_X(0.4)\approx0.7667.\]

Exercise 7: log-normal

The amount \(X\) spent on a type of repair is log-normal with \(\mu=\log(100)\) and \(\sigma=0.4\). Find the median, the mean, and \(\mathbb{P}(X>150)\).

Show solution

The median is \(e^\mu=100\). The mean is

\[e^{\mu+\sigma^2/2}=100e^{0.08}\approx108.33.\]

Also,

\[\mathbb{P}(X>150) =1-\Phi\!\left(\frac{\log(150)-\log(100)}{0.4}\right) \approx0.155.\]

Chapter summary

Uniform models bounded equal-density uncertainty; normal models symmetric measurements and approximates many sums; log-normal models positive quantities generated by multiplicative effects; exponential models constant-hazard waiting; gamma generalizes positive waiting and severity shapes; and beta models proportions. The support and generating mechanism should guide model selection, while interactive density and cdf comparisons reveal parameter effects.