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Introduction

Suppose that a random voltage in the range[0,1) is applied to a voltmeter with a one-digit display. Then the display output can be modeled by a discrete random variable Y taking values.0,.1,.2,...,.9 with P(Y = k/10) = 1/10 for k = 0,...,9. If this same random voltage is applied to a voltmeter with a two-digit display, then we can model its display output by a discrete random variable Z taking values.00,.01,...,.99 with P(Z = k/100) = 1/100 for k = 0,...,99. But how can we model the voltage itself? The voltage itself, call it X , can be any number in range[0,1). For example, if 0.15 ≤ X < 0.25, the one-digit voltmeter would round to the tens place and show Y = 0.2. In other words, we want to be able to write

P k

10− 0.05 ≤ X < k

10+ 0.05

= P Y= k

10

 = 1 10. Notice that 1/10 is the length of the interval

 k

10− 0.05, k

10+ 0.05 .

This suggests that probabilities involving X can be computed via P(a ≤ X < b) = ^b

a 1 dx = b − a,

which is the length of the interval[a,b). This observation motivates the concept of a con-tinuous random variable.

138

Definition

We say that X is a continuous random variable ifP(X ∈ B) has the form

P(X ∈ B) = 

B

f(t)dt :=

 _∞

−∞IB(t) f (t)dt (4.1) for some integrable function f .^aSinceP(X ∈ IR) = 1, the function f must integrate to one;

i.e.,_−∞^∞ f(t)dt = 1. Further, since P(X ∈ B) ≥ 0 for all B, it can be shown that f must be nonnegative.¹ A nonnegative function that integrates to one is called a probability density function (pdf).

Usually, the set B is an interval such as B= [a,b]. In this case, P(a ≤ X ≤ b) =

 _b a

f(t)dt.

See Figure 4.1(a). Computing such probabilities is analogous to determining the mass of a piece of wire stretching from a to b by integrating its mass density per unit length from a to b. Since most probability densities we work with are continuous, for a small interval, say [x,x + ∆x], we have

P(x ≤ X ≤ x + ∆x) = ^x+∆x

x f(t)dt ≈ f (x)∆x.

See Figure 4.1(b).

(a) (b)

a b x x+ x

Figure 4.1. (a) P(a ≤ X ≤ b) =a^bf(t)dt is the area of the shaded region under the density f (t). (b) P(x ≤ X ≤ x+ ∆x) =x^x+∆xf(t)dt is the area of the shaded vertical strip.

Note that for random variables with a density,

P(a ≤ X ≤ b) = P(a < X ≤ b) = P(a ≤ X < b) = P(a < X < b)

since the corresponding integrals over an interval are not affected by whether or not the endpoints are included or excluded.

Some common densities

Here are some examples of continuous random variables. A summary of the more com-mon ones can be found on the inside of the back cover.

aLater, when more than one random variable is involved, we write fX(x) instead of f (x).

Uniform. The simplest continuous random variable is the uniform. It is used to model experiments in which the outcome is constrained to lie in a known interval, say[a,b], and all outcomes are equally likely. We write f∼ uniform[a,b] if a < b and

f(x) =

This density is shown in Figure 4.2. To verify that f integrates to one, first note that since

a b _____1

b−a

Figure 4.2. The uniform density on [a,b].

f(x) = 0 for x < a and x > b, we can write

This calculation illustrates an important technique that is often incorrectly carried out by novice students: First modify the limits of integration, then substitute the appropriate for-mula for f(x). For example, it is quite common to see the incorrect calculation,

 _∞

−∞f(x)dx = ^∞

−∞

1

b− adx = ∞.

Example 4.1. In coherent radio communications, the phase difference between the trans-mitter and the receiver, denoted byΘ, is modeled as having a density f ∼ uniform[−π,π].

FindP(Θ ≤ 0) and P(Θ ≤π/2). restricted the limits of integration to be inside the region where the density is positive, we can write

P(Θ ≤ 0) = ⁰

−π

1

2π^dθ = 1/2.

The second probability is treated in the same way. First write P(Θ ≤π/2) = ^π/2

−∞ f(θ)dθ = ^π/2

−π f(θ)dθ.

It then follows that

P(Θ ≤π/2) = ^π/2

−π

1

2π^dθ = 3/4.

Example 4.2. Use the results of the preceding example to compute P(Θ >π/2|Θ > 0).

Solution. To calculate

P(Θ >π/2|Θ > 0) = P({Θ >π/2} ∩ {Θ > 0}) P(Θ > 0) ,

first observe that the denominator is simplyP(Θ > 0) = 1−P(Θ ≤ 0) = 1−1/2 = 1/2. As for the numerator, note that

{Θ >π/2} ⊂ {Θ > 0}.

Then use the fact that A⊂ B implies A ∩ B = A to write

P({Θ >π/2} ∩ {Θ > 0}) = P(Θ >π/2) = 1 − P(Θ ≤π/2) = 1/4.

Thus,P(Θ >π/2|Θ > 0) = (1/4)/(1/2) = 1/2.

Exponential.Another simple continuous random variable is the exponential with para-meterλ> 0. We write f ∼ exp(λ) if

f(x) =

+λ^e−λx, x ≥ 0, 0, x< 0.

This density is shown in Figure 4.3. Asλ increases, the height increases and the width decreases. It is easy to check that f integrates to one. The exponential random variable is often used to model lifetimes, such as how long a cell-phone call lasts or how long it takes a computer network to transmit a message from one node to another. The exponential random variable also arises as a function of other random variables. For example, in Problem 4.3 you will show that if U ∼ uniform(0,1), then X = ln(1/U) is exp(1). We also point out that if U and V are independent Gaussian random variables, which are defined later in this section, then U²+V²is exponential and√

U²+V²is Rayleigh (defined in Problem 30).² Example 4.3. Given that a cell-phone call has lasted more than t seconds so far, suppose the conditional probability that the call ends by t+∆t is approximatelyλ∆t when ∆t is small.

Show that the call duration is an exp(λ) random variable.

Solution. Let T denote the call duration. We treat the problem assumption as saying that

P(T ≤ t + ∆t|T > t) ≈ λ∆t.

exp(λ) λ

λ/2

0

0 ln(2)/λ

Laplace(λ) λ/2

λ/4

0

−ln(2)/λ0 ln(2)/λ

Cauchy(λ)

0

−λ 0 λ

N(m,σ²)

0 m−σ m m+σ

√2π σ__

_____1

2πλ πλ

1

_____1

____

Figure 4.3. Several common density functions.

To find the density of T , we proceed as follows. Let t≥ 0 and write P(T ≤ t + ∆t|T > t) = P({T ≤ t + ∆t} ∩ {T > t})

P(T > t)

= P(t < T ≤ t + ∆t) P(T > t)

=  _t+∆t

t fT(θ)dθ P(T > t) .

For small∆t, the left-hand side is approximatelyλ∆t, and the right-hand side is approxi-mately fT(t)∆t/P(T > t); i.e.,

λ∆t = fT(t)∆t P(T > t).

Now cancel∆t on both sides and multiply both sides by P(T > t) to getλP(T > t) = fT(t).

In this equation, writeP(T > t) as an integral to obtain λ^∞

t

fT(θ)dθ = fT(t).

Differentiating both sides with respect to t shows that

−λ^fT(t) = fT^(t), t ≥ 0.

The solution of this differential equation is easily seen to be fT(t) = ce^−λtfor some constant c. However, since fT(t) is a density and since its integral from zero to infinity must be one, it follows that c=λ^.

Remark. In the preceding example, T was the duration of a cell-phone call. However, if T were the lifetime or time-to-failure of a device or system, then in reliability theory, the quantity

∆t→0lim

P(T ≤ t + ∆t|T > t)

∆t

is called the failure rate. If this limit does not depend on t, then the calculation of the preceding example shows that the density of T must be exponential. Time-varying failure rates are considered in Section 5.7.

Laplace / double-sided exponential.Related to the exponential is the Laplace, some-times called the double-sided exponential. Forλ> 0, we write f ∼ Laplace(λ) if

f(x) = ^λ₂e^−λ|x|.

This density is shown in Figure 4.3. Asλ increases, the height increases and the width decreases. You will show in Problem 54 that the difference of two independent exp(λ) random variables is a Laplace(λ) random variable.

Example 4.4. An Internet router can send packets via route 1 or route 2. The packet delays on each route are independent exp(λ) random variables, and so the difference in delay between route 1 and route 2, denoted by X , has a Laplace(λ) density. Find

P(−3 ≤ X ≤ −2 or 0 ≤ X ≤ 3).

Solution. The desired probability can be written as

P({−3 ≤ X ≤ −2} ∪ {0 ≤ X ≤ 3}).

Since these are disjoint events, the probability of the union is the sum of the individual probabilities. We therefore need to compute

P(−3 ≤ X ≤ −2) and P(0 ≤ X ≤ 3).

Since X has a Laplace(λ) density, these probabilities are equal to the areas of the corre-sponding shaded regions in Figure 4.4. We first compute

P(−3 ≤ X ≤ −2) = ⁻²

−3

λ2e^−λ|x|dx = ^λ₂⁻²

−3 e^λxdx,

y/2

0 1 2 3

−1

−2

−3

Figure 4.4. Laplace(λ) density for Example 4.4.

where we have used the fact that since x is negative in the range of integration,|x| = −x.

This last integral is equal to(e^−2λ− e^−3λ)/2. It remains to compute P(0 ≤ X ≤ 3) = ³

0

λ2e^−λ|x|dx = ^λ₂³

0 e^−λxdx, which is equal to(1 − e^−3λ)/2. The desired probability is then

1− 2e^−3λ+ e^−2λ

2 .

Cauchy.The Cauchy random variable with parameterλ> 0 is also easy to work with.

We write f∼ Cauchy(λ) if

f(x) = λ/π λ²+ x².

This density is shown in Figure 4.3. Asλ increases, the height decreases and the width increases. Since (1/π)(d/dx)tan⁻¹(x/λ) = f (x), and since tan⁻¹(∞) =π/2, it is easy to check that f integrates to one. The Cauchy random variable arises as the tangent of a uniform random variable (Example 5.10) and also as the quotient of independent Gaussian random variables (Problem 33 in Chapter 7).

Example 4.5. In theλ-lottery you choose a numberλ ^{with 1}≤λ ≤ 10. Then a random variable X is chosen according to the Cauchy density with parameterλ^{. If}|X| ≥ 1, then you win the lottery. Which value ofλ should you choose to maximize your probability of winning?

Solution. Your probability of winning is

P(|X| ≥ 1) = P(X ≥ 1 or X ≤ −1)

= ^∞

1 f(x)dx +⁻¹

−∞ f(x)dx,

where f(x) = (λ/π)/(λ²+ x²) is the Cauchy density. Since the Cauchy density is an even function,

P(|X| ≥ 1) = 2  _∞

1

λ/π λ²+ x²dx. Now make the change of variable y= x/λ^{, dy}= dx/λ^{, to get}

P(|X| ≥ 1) = 2  _∞

1/λ

1/π 1+ y²dy.

Since the integrand is nonnegative, the integral is maximized by minimizing 1/λ ^{or by} maximizingλ. Hence, choosingλ= 10 maximizes your probability of winning.

Gaussian / normal.The most important density is the Gaussian or normal. Forσ²> 0, and convex for x outside this interval. Asσ increases, the height of the density decreases and it becomes wider as illustrated in Figure 4.5. If m= 0 andσ²= 1, we say that f is a

Figure 4.5. N(m,σ²) densities with different values ofσ.

As a consequence of the central limit theorem, whose discussion is taken up in Chap-ter 5, the Gaussian density is a good approximation for computing probabilities involving a sum of many independent random variables; this is true whether the random variables are continuous or discrete! For example, let

X := X1+ ··· + Xn,

where the Xi are i.i.d. with common mean m and common varianceσ². For large n, it is shown in Chapter 5 that if the Xiare continuous random variables, then

fX(x) ≈ √ 1

while if the Xiare integer-valued, pX(k) ≈ 1

In particular, since the macroscopic noise current measured in a circuit results from the sum of forces of many independent collisions on an atomic scale, noise current is well-described by the Gaussian density. For this reason, Gaussian random variables are the noise model of choice in electronic communication and control systems.

To verify that an arbitrary normal density integrates to one, we proceed as follows.

(For an alternative derivation, see Problem 17.) First, making the change of variable t= (x − m)/σ^{shows that} 

So, without loss of generality, we may assume f is a standard normal density with m= 0 Now write the product of integrals as the iterated integral

I² = ^∞

−∞

 _∞

−∞e^−(x2+y2)/2dx dy.

Next, we interpret this as a double integral over the whole plane and change from Cartesian coordinates x and y to polar coordinates r andθ. To integrate over the whole plane in polar coordinates, the radius r ranges from 0 to∞, and the angleθ ranges from 0 to 2π^{. The}

Example 4.6. The noise voltage in a certain amplifier has the standard normal density.

Show that the noise is as likely to be positive as it is to be negative.

Solution. In terms of the density, which we denote by f , we must show that  ₀

−∞f(x)dx = ^∞

0

f(x)dx.

Since f(x) = e^−x2/2/√

2πis an even function of x, the two integrals are equal. Furthermore, we point out that since the sum of the two integrals is _−∞^∞ f(x)dx = 1, each individual integral must be 1/2.

Location and scale parameters and the gamma densities

Since a probability density function can be any nonnegative function that integrates to one, it is easy to create a whole family of density functions starting with just one density function. Let f be any nonnegative function that integrates to one. For any real number c and any positive numberλ, consider the nonnegative function

λ ^fλ(x − c) .

Here c is called a location parameter andλ is called a scale parameter. To show that this new function is a probability density, all we have to do is show that it integrates to one. In

the integral 

∞

−∞λ ^fλ(x − c) dx

0 1 example, if f is the triangular density shown in Figure 4.6(a), then f(x − c) is the density shown in Figure 4.6(b). If c is positive, then f(x − c) is f (x) shifted to the right, and if c is negative, then f(x − c) is f (x) shifted to the left.

Next consider the case c= 0 andλ> 0. In this case, the main effect ofλ is to shrink (if λ> 1) or to expand (ifλ< 1) the density. The second effect ofλ is to increase or decrease the height of the density. For example, if f is again the triangular density of Figure 4.6(a), thenλ^f(λ^x) is shown in Figure 4.6(c) for 0 <λ < 1.

To see what happens both c= 0 andλ> 0, first put h(x) :=λ ^f(λ^x). Then observe that h(x − c) =λ ^fλ(x − c)

. In other words, first find the picture forλ^f(λ^x), and then shift this picture by c.

In the exponential and Laplace densities,λ is a scale parameter, while in the Cauchy density, 1/λ is a scale parameter. In the Gaussian, if we write f(x) = e^−x2/2/√ random variable, m is a location parameter and 1/σ is a scale parameter. Note in particular that asσincreases, the density becomes shorter and wider, while asσdecreases, the density becomes taller and narrower (recall Figure 4.5).

An important application of the scale parameter arises with the basic gamma density with parameter p> 0. This density is given by

gp(x) :=

is the gamma function. In other words, the gamma function is defined to make the gamma density integrate to one.³ (Properties of the gamma function are derived in Problem 14.) Graphs of gp(x) for p = 1/2, p = 1, p = 3/2, p = 2, and p = 3 are shown in Figure 4.7. To explain the shapes of these curves, observe that for x near zero, e^−x≈ 1, and so the behavior is determined by the factor x^p−1. For the values of p in the figure, this factor is x^−1/2, x⁰, x^1/2, x, and x². In the first case, x^−1/2blows up as x approaches the origin, and decreases as x moves to the right. Of course, x⁰= 1 is a constant, and in the remaining cases, x^p−1 is zero for x= 0 and then increases. In all cases, as x moves to the right, eventually, the decaying nature of the factor e^−xdominates, and the curve decreases to zero as x→ ∞.

Setting gp,λ(x) :=λ^gp(λx) defines the general gamma density, and the following spe-cial cases are of great importance. When p= m is a positive integer, gm,λ is called an Erlang(m,λ) density (see Problem 15). As shown in Problem 55(c), the sum of m i.i.d.

exp(λ) random variables is an Erlang(m,λ) random variable. For example, if m customers are waiting in a queue, and the service time for each one is exp(λ), then the time to serve all m is Erlang(m,λ). The Erlang densities for m = 1,2,3 andλ= 1 are g1(x), g2(x), and g3(x) shown in Figure 4.7. When p= k/2 andλ= 1/2, g_p,λ is called a chi-squared density with k degrees of freedom. As you will see in Problem 46, the chi-squared random variable arises as the square of a normal random variable. In communication systems employing nonco-herent receivers, the incoming signal is squared before further processing. Since the thermal noise in these receivers is Gaussian, chi-squared random variables naturally appear. Since chi-squared densities are scaled versions of g_k/2, the chi-squared densities for k= 1,2,3,4, and 6 are scaled versions of g_1/2, g1, g_3/2, g2, and g3shown in Figure 4.7.

0 1 2 3 4

0 1

= 1/2

= 1

= 3/2 p

p

p

p = 2

p = 3

Figure 4.7. The gamma densities gp(x) for p = 1/2, p = 1, p = 3/2, p = 2, and p = 3.

The paradox of continuous random variables

Let X be a continuous random variable. For any given x0, write 1 = ^∞

−∞f(t)dt = ^x0

−∞f(t)dt +^∞

x₀ f(t)dt

= P(X ≤ x0) + P(X ≥ x0)

= P(X ≤ x0) + P(X = x0) + P(X > x0).

SinceP(X ≤ x0) + P(X > x0) = P(X ∈ IR) = 1, it follows that P(X = x0) = 0. We are thus confronted with the fact that continuous random variables take no fixed value with positive probability! The way to understand this apparent paradox is to realize that continuous ran-dom variables are an idealized model of what we normally think of as continuous-valued measurements. For example, a voltmeter only shows a certain number of digits after the decimal point, say 5.127 volts because physical devices have limited precision. Hence, the measurement X= 5.127 should be understood as saying that

5.1265 ≤ X < 5.1275,

since all numbers in this range round to 5.127. Now there is no paradox since P(5.1265 ≤ X< 5.1275) has positive probability.

You may still ask, “Why not just use a discrete random variable taking the distinct val-ues k/1000, where k is any integer?” After all, this would model the voltmeter in qval-uestion.

One answer is that if you get a better voltmeter, you need to redefine the random variable, while with the idealized, continuous-random-variable model, even if the voltmeter changes, the random variable does not. Also, the continuous-random-variable model is often mathe-matically simpler to work with.

Remark. If B is any set with finitely many points, or even countably many points, then P(X ∈ B) = 0 when X is a continuous random variable. To see this, suppose B = {x1,x2,...}

where the xiare distinct real numbers. Then P(X ∈ B) = P

_∞

i=1

{xi}



=

∑

^∞

i=1P(X = xi) = 0, since, as argued above, each term is zero.

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