14.1 - Probability Density Functions

A continuous random variable takes on an uncountably infinite number of possible values. For a discrete random variable \(X\) that takes on a finite or countably infinite number of possible values, we determined \(P(Ten=x)\) for all of the possible values of \(X\), and called information technology the probability mass part ("p.m.f."). For continuous random variables, as we shall soon see, the probability that \(X\) takes on any detail value \(10\) is 0. That is, finding \(P(X=x)\) for a continuous random variable \(10\) is not going to work. Instead, nosotros'll need to find the probability that \(X\) falls in some interval \((a, b)\), that is, we'll need to find \(P(a<X<b)\). Nosotros'll practise that using a probability density function ("p.d.f."). We'll showtime motivate a p.d.f. with an example, and so we'll formally ascertain it.

Example 14-1 Section

quarter pounder burger

Fifty-fifty though a fast-food concatenation might advertise a hamburger as weighing a quarter-pound, you tin well imagine that it is non exactly 0.25 pounds. 1 randomly selected hamburger might weigh 0.23 pounds while another might weigh 0.27 pounds. What is the probability that a randomly selected hamburger weighs between 0.20 and 0.30 pounds? That is, if we let \(X\) denote the weight of a randomly selected quarter-pound hamburger in pounds, what is \(P(0.twenty<X<0.30)\)?

Solution

In reality, I'm not particularly interested in using this example simply then that you'll know whether or non yous've been ripped off the adjacent time you lodge a hamburger! Instead, I'k interested in using the instance to illustrate the thought behind a probability density function.

Now, you could imagine randomly selecting, allow'due south say, 100 hamburgers advertised to weigh a quarter-pound. If you weighed the 100 hamburgers, and created a density histogram of the resulting weights, possibly the histogram might look something like this:

X 0.25 Density

In this case, the histogram illustrates that about of the sampled hamburgers exercise indeed weigh close to 0.25 pounds, but some are a bit more and some a bit less. At present, what if nosotros decreased the length of the class interval on that density histogram? And so, the density histogram would expect something like this:

X 0.25 Density

Now, what if we pushed this further and decreased the intervals fifty-fifty more than? You lot tin can imagine that the intervals would eventually get so small that nosotros could represent the probability distribution of \(X\), not as a density histogram, but rather as a curve (by connecting the "dots" at the tops of the tiny tiny tiny rectangles) that, in this instance, might wait like this:

X 0.25 f(10)

Such a curve is denoted \(f(10)\) and is called a (continuous) probability density function.

Now, you might retrieve that a density histogram is defined so that the area of each rectangle equals the relative frequency of the corresponding class, and the area of the unabridged histogram equals 1. That suggests then that finding the probability that a continuous random variable \(X\) falls in some interval of values involves finding the area under the curve \(f(10)\) sandwiched past the endpoints of the interval. In the case of this example, the probability that a randomly selected hamburger weighs betwixt 0.20 and 0.30 pounds is then this area:

Ten 0.20 0.xxx f(x) Area = Probability P(0.20<X<0.30)

Now that we've motivated the idea backside a probability density function for a continuous random variable, let's now go and formally define information technology.

Probability Density Office ("p.d.f.")

The probability density function ("p.d.f.") of a continuous random variable \(10\) with support \(Due south\) is an integrable function \(f(x)\) satisfying the post-obit:

  1. \(f(x)\) is positive everywhere in the back up \(S\), that is, \(f(x)>0\), for all \(x\) in \(South\)

  2. The area under the curve \(f(x)\) in the support \(S\) is i, that is:

    \(\int_S f(x)dx=1\)

  3. If \(f(10)\) is the p.d.f. of \(x\), then the probability that \(10\) belongs to \(A\), where \(A\) is some interval, is given by the integral of \(f(x)\) over that interval, that is:

    \(P(X \in A)=\int_A f(x)dx\)

As yous can see, the definition for the p.d.f. of a continuous random variable differs from the definition for the p.1000.f. of a detached random variable by simply irresolute the summations that appeared in the discrete instance to integrals in the continuous case. Allow's examination this definition out on an case.

Example xiv-2 Section

Allow \(X\) be a continuous random variable whose probability density function is:

\(f(x)=3x^2, \qquad 0<x<1\)

First, annotation once again that \(f(ten)\ne P(X=ten)\). For example, \(f(0.9)=3(0.9)^2=ii.43\), which is clearly non a probability! In the continuous example, \(f(ten)\) is instead the height of the bend at \(X=x\), so that the total surface area under the bend is 1. In the continuous case, information technology is areas under the curve that ascertain the probabilities.

Now, permit's commencement start by verifying that \(f(x)\) is a valid probability density function.

Solution

What is the probability that \(X\) falls between \(\frac{1}{2}\) and one? That is, what is \(P\left(\frac{1}{2}<X<1\right)\)?

Solution

What is \(P\left(X=\frac{1}{ii}\right)\)?

Solution

Information technology is a straightforward integration to encounter that the probability is 0:

\(\int^{1/2}_{1/2} 3x^2dx=\left[10^3\right]^{ten=1/2}_{ten=1/2}=\dfrac{1}{viii}-\dfrac{1}{8}=0\)

In fact, in general, if \(10\) is continuous, the probability that \(X\) takes on whatever specific value \(10\) is 0. That is, when \(X\) is continuous, \(P(Ten=x)=0\) for all \(x\) in the support.

An implication of the fact that \(P(Ten=x)=0\) for all \(x\) when \(Ten\) is continuous is that you can be careless almost the endpoints of intervals when finding probabilities of continuous random variables. That is:

\(P(a\le Ten\le b)=P(a<X\le b)=P(a\le Ten<b)=P(a<ten<b)\)

for any constants \(a\) and \(b\).

Instance 14-iii Section

Let \(X\) be a continuous random variable whose probability density function is:

\(f(x)=\dfrac{x^3}{4}\)

for an interval \(0<ten<c\). What is the value of the constant \(c\) that makes \(f(x)\) a valid probability density function?

Solution