Unit 10, lesson 67

Choosing a Continuous Probability Model

How to choose a uniform, exponential or normal distribution family from the mechanism, the support and the tail behavior, why the parameter convention matters, and how model adequacy differs from a correct calculation.

  • Video 67
  • Choosing a Continuous Model

Two constructed models for a wait both have a mean of 55 seconds. One is uniform on [0,10][0, 10] and the other is exponential, and at 88 seconds their survival probabilities nearly agree. Yet the uniform model gives a wait longer than 1010 seconds probability 00, while the exponential model gives it probability about 0.1350.135. A curve that looks plausible cannot decide between them. This lesson chooses among uniform, exponential and normal models from the mechanism behind a quantity, the values it can take, the meaning of the parameters and the behavior of the tails, and it separates a correct calculation inside a model from the question of whether the model fits.

Key terms

Distribution family

Definition A distribution family is a set of distributions described by one formula with parameters, where each allowed choice of parameter values gives one member. The continuous families compared in this lesson are the uniform laws Uniform⁡(a,b)\operatorname{Uniform}(a, b) with a<ba < b, the exponential laws Exp⁡(λ)\operatorname{Exp}(\lambda) with λ>0\lambda > 0, and the normal laws N(μ,σ2)N(\mu, \sigma^2) with real μ\mu and σ>0\sigma > 0. Choosing a family is an assumption about how the quantity arises, and choosing the parameter values then picks one member. Each family rests on its own assumption.

  • A uniform law on [a,b][a, b] gives every subinterval (c,d)(c, d) of [a,b][a, b] the probability (d−c)/(b−a)(d - c)/(b - a), so subintervals of equal length are equally likely.
  • An exponential law is memoryless. A continuous random variable TT with values in [0,∞)[0, \infty) and P(T>t)>0P(T > t) > 0 for every t≥0t \ge 0 satisfies P(T>t+s∣T>t)=P(T>s)P(T > t + s \mid T > t) = P(T > s) for all s,t≥0s, t \ge 0 exactly when T∼Exp⁡(λ)T \sim \operatorname{Exp}(\lambda) for some λ>0\lambda > 0.
  • A normal law is symmetric about its mean, gives positive probability to every interval of positive length and has very thin tails. It is a common model for a signed measurement error that is roughly symmetric, although symmetry alone does not make a distribution normal.

In plain words Pick the family from what is known about how the quantity is produced, then set the parameters. The look of a curve is weak evidence. Knowing only that a quantity lies between 00 and 1010 does not make it uniform, since a process on that range may favor the middle, the ends or one side. A matching mean does not identify the family either.

Example Take three constructed situations, all measured in seconds. A timing offset UU is generated uniformly between 00 and 1010, a wait TT is exponential with mean 55, and a signed error EE is normal with mean 00 and standard deviation 22. These are illustrative assumptions, not measurements of a real system. In this course’s notation

U∼Uniform⁡(0,10),T∼Exp⁡(0.2),E∼N(0,4).\begin{gathered} U \sim \operatorname{Uniform}(0, 10), \\[0.4em] T \sim \operatorname{Exp}(0.2), \\[0.4em] E \sim N(0, 4). \end{gathered}

The generation rule justifies uniformity, so P(2<U<5)=3/10=0.3P(2 < U < 5) = 3/10 = 0.3, the length of the interval divided by the length of the support. Both UU and TT have mean 55 seconds, but

Var⁡(U)=10212=253≈8.33,Var⁡(T)=10.22=25,\begin{aligned} \operatorname{Var}(U) &= \frac{10^2}{12} = \frac{25}{3} \\[0.4em] &\approx 8.33, \\[0.4em] \operatorname{Var}(T) &= \frac{1}{0.2^2} = 25, \end{aligned}

both in seconds squared. The same mean fits two different families, so the mean alone cannot choose between them.

Parameter convention

Definition A parameter convention states what each parameter of a family means, which values it may take and in which units it is measured. This course uses the following conventions.

  • The uniform law Uniform⁡(a,b)\operatorname{Uniform}(a, b) is written with its endpoints a<ba < b. Its density is 1/(b−a)1/(b - a) between aa and bb and 00 outside [a,b][a, b], its mean is (a+b)/2(a + b)/2 and its variance is (b−a)2/12(b - a)^2/12.
  • The exponential law Exp⁡(λ)\operatorname{Exp}(\lambda) is written with its rate λ>0\lambda > 0, measured per unit of time. Its density is f(t)=λe−λtf(t) = \lambda e^{-\lambda t} for t≥0t \ge 0 and 00 for t<0t < 0, its survival probability is P(T>t)=e−λtP(T > t) = e^{-\lambda t} for t≥0t \ge 0, its mean is 1/λ1/\lambda and its variance is 1/λ21/\lambda^2.
  • The normal law N(μ,σ2)N(\mu, \sigma^2) is written with its mean μ\mu and its variance σ2>0\sigma^2 > 0, so its standard deviation is σ\sigma.

Other books write the exponential law with its mean θ=1/λ\theta = 1/\lambda, the scale parameter, or put the standard deviation of a normal law in the second place. If a random variable is multiplied by a constant c>0c > 0, as in a change of units, then

cU∼Uniform⁡(ca,cb),cT∼Exp⁡(λ/c),cX∼N(cμ,c2σ2),\begin{gathered} cU \sim \operatorname{Uniform}(ca, cb), \\[0.4em] cT \sim \operatorname{Exp}(\lambda / c), \\[0.4em] cX \sim N(c\mu, c^2\sigma^2), \end{gathered}

where U∼Uniform⁡(a,b)U \sim \operatorname{Uniform}(a, b), T∼Exp⁡(λ)T \sim \operatorname{Exp}(\lambda) and X∼N(μ,σ2)X \sim N(\mu, \sigma^2). An event that describes the same physical outcome in the new units has the same probability.

In plain words Before comparing formulas or answers, write down which number is a rate and which is a mean, whether the second normal parameter is a variance or a standard deviation, and the unit of each. Endpoints, means and standard deviations scale with the quantity, a variance scales with the square of the factor, and a rate scales the opposite way, because it counts per unit of time.

Example The wait TT has mean 55 seconds, so its rate is λ=1/5=0.2\lambda = 1/5 = 0.2 per second, which is 0.2⋅60=120.2 \cdot 60 = 12 per minute. Eight seconds is 8/60=2/158/60 = 2/15 of a minute, and rate times time is the same in both units,

0.2⋅8=12⋅215=1.6,0.2 \cdot 8 = 12 \cdot \tfrac{2}{15} = 1.6,

so the probability of waiting more than 88 seconds is e−1.6≈0.2019e^{-1.6} \approx 0.2019 either way. Keeping the rate 0.20.2 while measuring time in minutes would give e−0.2⋅2/15≈0.9737e^{-0.2 \cdot 2/15} \approx 0.9737, a sign of a unit mistake. Writing Exp⁡(5)\operatorname{Exp}(5) for this wait would mean, in this course’s convention, a rate of 55 per second and a mean of 0.20.2 seconds. In the same way, E∼N(0,4)E \sim N(0, 4) has standard deviation 22 seconds, so P(∣E∣>4)=2(1−Φ(2))≈0.0455P(\lvert E \rvert > 4) = 2\bigl(1 - \Phi(2)\bigr) \approx 0.0455. Reading the 44 as a standard deviation would give 2(1−Φ(1))≈0.31732\bigl(1 - \Phi(1)\bigr) \approx 0.3173 instead.

Support of a model

Definition The support of a distribution is the set of values xx such that every interval (x−ε,x+ε)(x - \varepsilon, x + \varepsilon) with ε>0\varepsilon > 0 has positive probability. It is the smallest closed set with probability 11. The supports of these three families are [a,b][a, b] for Uniform⁡(a,b)\operatorname{Uniform}(a, b), the half-line [0,∞)[0, \infty) for Exp⁡(λ)\operatorname{Exp}(\lambda) and the whole real line for N(μ,σ2)N(\mu, \sigma^2). Under a continuous model every single value has probability 00, including the values in the support, so the support is a statement about intervals rather than single points. A model that gives positive probability to an interval of values the quantity cannot take, or probability 00 to an interval the quantity reaches with positive probability, is not an exact description. At best it is an approximation whose error has to be judged.

In plain words The support says where the model allows values. Compare it with the values the quantity can really take before computing anything. A quantity confined to a bounded interval, a nonnegative quantity with no upper limit and a signed quantity with no fixed limits call for different supports.

Example The offset UU has support [0,10][0, 10], the wait TT has support [0,∞)[0, \infty) and the error EE has support the whole real line. The value P(U>10)=0P(U > 10) = 0 is right for an offset generated inside [0,10][0, 10], but a uniform model on [0,10][0, 10] for a wait that can exceed 1010 seconds would wrongly give that event probability 00. Now suppose someone proposes W∼N(5,25)W \sim N(5, 25), with mean 55 seconds and standard deviation 55 seconds, for a wait that cannot be negative. Then

P(W<0)=Φ ⁣(0−55)=Φ(−1)≈0.1587,\begin{aligned} P(W < 0) &= \Phi\!\left(\frac{0 - 5}{5}\right) \\[0.4em] &= \Phi(-1) \approx 0.1587, \end{aligned}

so the model puts nearly 1616 percent of its probability on impossible waits. The exponential model gives P(T<0)=0P(T < 0) = 0.

Tail behavior

Definition The tail behavior of a distribution is the way its tail probabilities P(X>x)P(X > x) and P(X<−x)P(X < -x) shrink as xx grows. One way to compare tails is the conditional probability P(X>x+s∣X>x)P(X > x + s \mid X > x) of going a further distance s>0s > 0 past a level xx already passed, defined when P(X>x)>0P(X > x) > 0. The three families behave differently.

  • For X∼Uniform⁡(a,b)X \sim \operatorname{Uniform}(a, b) the tail probabilities are exactly 00 above bb and below aa. For a≤x<ba \le x < b the conditional probability equals 1−s/(b−x)1 - s/(b - x) when x+s<bx + s < b and 00 otherwise, so it falls as xx approaches bb.

  • For T∼Exp⁡(λ)T \sim \operatorname{Exp}(\lambda) the conditional probability does not depend on the level. For every x≥0x \ge 0 and s>0s > 0,

    P(T>x+s∣T>x)=e−λ(x+s)e−λx=e−λs.\begin{aligned} &P(T > x + s \mid T > x) \\[0.4em] &\quad = \frac{e^{-\lambda(x + s)}}{e^{-\lambda x}} = e^{-\lambda s}. \end{aligned}

    This is the memoryless property.

  • For X∼N(μ,σ2)X \sim N(\mu, \sigma^2) the conditional probability tends to 00 as x→∞x \to \infty, for every fixed s>0s > 0, and the left tail behaves the same way by symmetry. Far out, a normal tail shrinks faster than any exponential tail.

Models that agree near the center can differ greatly in the tails, so tail behavior matters most when the question is about rare or extreme values.

In plain words The tails describe how a model treats large values. A bounded model rules out values past its endpoint, and the chance of going a further distance drops to 00 as the endpoint nears. An exponential wait has the same chance of lasting ss more seconds however long it has already lasted. A normal model gives positive probability to every interval of positive length, however far out, but the chance of going a further fixed distance keeps shrinking the farther out the level is.

Example At 88 seconds the uniform and exponential survival probabilities nearly agree, with P(U>8)=0.2P(U > 8) = 0.2 and P(T>8)=e−1.6≈0.2019P(T > 8) = e^{-1.6} \approx 0.2019. At 1010 seconds they do not, since P(U>10)=0P(U > 10) = 0 while P(T>10)=e−2≈0.1353P(T > 10) = e^{-2} \approx 0.1353. Given that the offset exceeds 88, it is uniform on (8,10)(8, 10), so P(U>9∣U>8)=1/2P(U > 9 \mid U > 8) = 1/2. Memorylessness gives

P(T>9∣T>8)=e−0.2≈0.8187\begin{aligned} P(T > 9 \mid T > 8) &= e^{-0.2} \\[0.4em] &\approx 0.8187 \end{aligned}

for the wait. For the normal error, P(E>4)=1−Φ(2)≈0.0228P(E > 4) = 1 - \Phi(2) \approx 0.0228, P(E>6)=1−Φ(3)≈0.00135P(E > 6) = 1 - \Phi(3) \approx 0.00135 and P(E>8)=1−Φ(4)≈0.0000317P(E > 8) = 1 - \Phi(4) \approx 0.0000317. Each further 22 seconds multiplies the normal tail by a factor that falls from about 0.0590.059 to about 0.0230.023, while the exponential tail is multiplied by the same factor e−0.4≈0.670e^{-0.4} \approx 0.670 over every 22 seconds.

Model adequacy

Definition Model adequacy is the judgment that a model’s answers are accurate enough for a stated purpose, especially for the events that purpose depends on. It is separate from mathematical correctness, which asks only whether the calculations follow from the stated model. A careful choice states the quantity and its units, the values it can take, the mechanism and the assumptions behind the family, and the parameters with their convention. It then checks the consequences that matter, such as the support, the tails and the probabilities a decision uses, against what is known. When data are available, the checks look at bounds, asymmetry, extreme values, changes over time and dependence between observations, not only at the average.

In plain words Calculations that follow correctly from a model describe the model. Whether they describe the real quantity depends on whether the model fits it, and correct algebra cannot settle that. Adequacy also depends on the use, since a model can describe the center well and still misjudge a rare extreme. Check the part of the distribution the question is about, report the assumptions, and revisit them when observations contradict their consequences.

Example The model W∼N(5,25)W \sim N(5, 25) for a nonnegative wait puts about 0.15870.1587 of its probability below 00, which is a serious defect for most uses. Now take a constructed nonnegative quantity with mean 5050 and standard deviation 55, modeled as N(50,25)N(50, 25). The model’s probability of a negative value is

Φ ⁣(0−505)=Φ(−10)≈7.6×10−24,\begin{aligned} \Phi\!\left(\frac{0 - 50}{5}\right) &= \Phi(-10) \\[0.4em] &\approx 7.6 \times 10^{-24}, \end{aligned}

so the impossible tail is negligible at any practical accuracy, although the other checks still apply. The same family can be adequate in one situation and not in another. Likewise, an exponential model for a wait with a known maximum is not exact, because it puts positive probability on the impossible waits beyond that maximum. It needs revision unless that probability is negligible for the purpose.

Check yourself

  1. A generator selects a real number VV uniformly from [−2,2][-2, 2], and a signed error ZZ is modeled as standard normal. For each model, state the support and find the probability that the magnitude is less than 11 and the probability that it exceeds 22.
Show the answer

The generation rule gives V∼Uniform⁡(−2,2)V \sim \operatorname{Uniform}(-2, 2), with support [−2,2][-2, 2] and density 14\tfrac14 between −2-2 and 22. The event ∣V∣<1\lvert V \rvert < 1 is the interval (−1,1)(-1, 1) of length 22, so its probability is 2/4=122/4 = \tfrac12. The event ∣V∣>2\lvert V \rvert > 2 lies outside the support, so its probability is 00.

The error Z∼N(0,1)Z \sim N(0, 1) has the whole real line as support, and

P(∣Z∣<1)=2Φ(1)−1≈0.6827,P(∣Z∣>2)=2(1−Φ(2))≈0.0455.\begin{aligned} P(\lvert Z \rvert < 1) &= 2\Phi(1) - 1 \\[0.4em] &\approx 0.6827, \\[0.4em] P(\lvert Z \rvert > 2) &= 2\bigl(1 - \Phi(2)\bigr) \\[0.4em] &\approx 0.0455. \end{aligned}

Both models are symmetric about 00, but the normal error can exceed 22 in magnitude and the generated number cannot.

  1. A waiting mechanism is memoryless, and the wait has mean 33 minutes. Choose a model, find the probability that the wait exceeds 66 minutes, and find the probability that it exceeds 66 minutes given that it has already exceeded 44 minutes. Would the model still fit if the mechanism guaranteed that no wait lasts more than 66 minutes?
Show the answer

Among continuous waits the memoryless ones are the exponential laws, and a mean of 33 minutes means a rate of 1/31/3 per minute, so S∼Exp⁡(1/3)S \sim \operatorname{Exp}(1/3). Then

P(S>6)=e−6/3=e−2≈0.1353.\begin{aligned} P(S > 6) &= e^{-6/3} = e^{-2} \\[0.4em] &\approx 0.1353. \end{aligned}

By memorylessness, P(S>6∣S>4)=P(S>2)=e−2/3≈0.5134P(S > 6 \mid S > 4) = P(S > 2) = e^{-2/3} \approx 0.5134. Working in seconds gives the same answers, since the rate is 1/1801/180 per second and 360/180=2360/180 = 2. Taking the rate to be 33 would confuse the rate with the mean and describe a wait with mean 2020 seconds.

With a strict maximum of 66 minutes the exponential model no longer fits. Its support [0,∞)[0, \infty) contains impossible waits, and it gives the impossible event S>6S > 6 probability about 0.13530.1353. In fact no memoryless model fits such a wait. A continuous wait with mean 33 and maximum 66 has P(S>3)>0P(S > 3) > 0, since otherwise it would equal 33 with probability 11, which a continuous wait cannot do. Memorylessness would then give P(S>6∣S>3)=P(S>3)P(S > 6 \mid S > 3) = P(S > 3), and multiplying by P(S>3)P(S > 3) gives P(S>6)=P(S>3)2>0P(S > 6) = P(S > 3)^2 > 0, which contradicts the maximum.

  1. Someone proposes replacing the exponential model of question 2 by the normal model W∼N(3,9)W \sim N(3, 9), which has the same mean and the same variance. Find P(W<0)P(W < 0), and compare the probability of a wait longer than 1212 minutes under the two models.
Show the answer

The normal model has mean 33 and standard deviation 33, so

P(W<0)=Φ(−1)≈0.1587.\begin{aligned} P(W < 0) &= \Phi(-1) \\[0.4em] &\approx 0.1587. \end{aligned}

It puts about 1616 percent of its probability on negative waits, which are impossible. For 1212 minutes the standardized value is (12−3)/3=3(12 - 3)/3 = 3, so

P(W>12)=1−Φ(3)≈0.00135,\begin{aligned} P(W > 12) &= 1 - \Phi(3) \\[0.4em] &\approx 0.00135, \end{aligned}

while the exponential model gives P(S>12)=e−4≈0.0183P(S > 12) = e^{-4} \approx 0.0183, about 13.613.6 times as large. Matching the mean and the variance does not match the support or the tail. If the mechanism really is memoryless, the normal model understates this tail probability by that factor.

How this lesson connects