The CDF and Conditional Probability

The CDF and Conditional Probability Brett Block, Anthony Schreck and Austin Smith November 15, 2010

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Conditional CDFs

We are used to playing with conditional distributions in terms of the PDF, but there is more than one way to skin a cat. Here we will work with conditional distributions in terms of the CDF. To begin, the general form of a conditional CDF is as follows1 : FXY (x | y) = P r[X ≤ x | Y ≤ y] =

FXY (x,y) FY (y)

Where FXY (x, y) denotes the joint CDF and FY (y) is the marginal CDF of the random variable Y . Please note, this general formula can be applied to continuous, discrete and mixture distributions. Dividing the joint CDF by the marginal CDF allows us to normalize our conditional distribution so that it maintains the necessary properties of the CDF only in terms of the variable we are not conditioning upon. For example, if we looked at the formula above, this particular conditional CDF only maintains the properties of the CDF in terms of X. Recall the properties of the CDF: 1) limx,y→−∞ FXY (x, y) = 0 and limx,y→∞ FXY (x, y) = 1 2)FXY (x, y) is a monotone, nondecreasing function for both X and Y 3)FXY (x, y) is continuous from the right for both X and Y Remember, the conditional CDF retains all the same properties as the CDF for only the random variable you are not conditioning upon. Later on, we will use an example to demonstrate that these properties hold for conditional CDFs.

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Kingsbury, Nick. Joint and Conditional cdfs and pdfs. Connexions. 7 June 2005 .

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1.1

Continuous Conditional CDFs

We will begin by looking at the case where the variables are jointly continuous. In terms of the CDF there is one primary method we use; it is direct, in so far that we go from a CDF to a conditional CDF. There is an alternative method that involves taking the conditional PDF and transforming it into a conditional CDF. We will focus on the first method, as it better helps us understand the conditional CDF as its own beast, independent of the PDF. 1.1.1

Direct CDF Method

Transforming a joint CDF into a conditional CDF requires that we figure out what the marginal CDF is. The marginal CDF of X can be found by taking the limit of the joint CDF as Y approaches infinity: FX (x) = limy→∞ FXY (x, y) And conversely for the marginal CDF of Y: FY (y) = limx→∞ FXY (x, y) Now that we know how to find the marginal CDF, we can use the following equation to determine the conditional CDF, FXY (x | y): FXY (x | y) = P r[X ≤ x | Y ≤ y] =

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FXY (x,y) FY (y)

1.1.2

Moments in the Conditional CDF

It may prove useful to obtain the first moment of a conditional CDF. That is, what is the expectation of a random variable X given that we know that the random variable Y is less than a given value. Notationally, we are interested in E[X | Y ≤ y]. To do this, first recall the formula for the mean of the CDF: E[X] =

R∞ 0

1 − FX (x)dx −

R0

−∞

FX (x)dx

So our desired expectation will follow a near identical form: E[X | Y ≤ y] =

R∞ 0

1−

FXY (x,y) dx FY (y)

−

R0

−∞

FXY (x,y) dx FY (y)

What you may notice here is that we are only integrating with respect to X. This leads us to an interesting point about this expectation: it will vary with Y . This makes intuitive sense, as you would expect that E[X | Y ≤ a] 6=E[X | Y ≤ b].

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1.1.3

Direct CDF Method Examples

Example 1: Taking a new spin on an example from the lecture on joint density functions, where we were given the following CDF:

FXY (x, y) =

0 if x < 0 or y < 0 .5xy(x + y) if 0 ≤ x < 1 and 0 ≤ y < 1 1 if x ≥ 1 and y ≥ 1 .5x(x + 1) if 0 ≤ x < 1 and y ≥ 1 .5y(1 + y) if x ≥ 1 and 0 ≤ y < 1

First we will generally solve for the conditional CDF, which we can then use to find our probabilities of interest. To solve this problem we need the joint CDF (which we have) and the marginal CDF (which we do not have, yet). Specifically, we need the marginal CDF of Y:

FY (y) = limx→∞ FXY (x, y) =

0 if y < 0 .5y(1 + y) if 0 ≤ y < 1 1 if y ≥ 1

Deriving the conditional CDF: FXY (x | y) =

FXY (x,y) FY (y)

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=

.5xy(x+y) .5y(1+y)

x + 0 y, x * y 1

( 12 + 12 y 2 ) 1 2 ( 2 x + 12 y 2 ) ( 21 x2 + 12 )

F (x, y) =

Use the conditional CDF to determine the probability that X is less than or equal to given Y is less than or equal to .25. Answer: We want to solve for P r(X ≤ .8 | Y ≤ .25): P r(X ≤ .8 | Y ≤ .25)=FX|Y (.8 | .25) = FXY (.8,.25) FY (.25)

=

(.5)(.82 )+(.5)(.252 ) (.5)+(.5)(.25)

=

.35125 .53125

FXY (.8,.25) FY (.25)

= .6611

Thus, our answer is that there is approximately a 66.11% chance that X is less than or equal to .8 given that Y is less than or equal to .25. 3) Given the CDF: 0 if X < 0 or Y < 0 + if X > 1 and 0 ≤ Y ≤ 1 ( 13 xy + 23 x) if 0 ≤ X ≤ 1 and 0 ≤ Y ≤ 1 ( 31 x + 23 x) if 0 ≤ X ≤ 1 and Y > 1 1 if X, Y > 1 ( 31 y

F (x, y) =

2 ) 3

What is P r[.5 ≤ X ≤ .9 | Y ≤ .4]? 18

Answer: Solve it as the difference between two conditional CDFs. P r[X ≤ .9 | Y ≤ .4] =

FXY (.9,.4) FY (.4)

=

(.5,.4) = P r[X ≤ .5 | Y ≤ .4] = FXY FY (.4)

( 13 )(.9)(.4)+( 23 )(.9) 1 (.4)+ 32 3 1 (.5)(.4)+ 23 (.5) 3 1 (.4)+ 23 3

=

= .4 .8

.72 .8

= .9

= .5

As P r[.5 ≤ X ≤ .9 | Y ≤ .4] = P r[X ≤ .9 | Y ≤ .4] − P r[X ≤ .5 | Y ≤ .4], we get .9 − .5 = .4 So there is a 40% chance that X is between .5 and .9, given that Y is less than or equal to .4.

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Works Cited

• Kingsbury, Nick. Joint and Conditional cdfs and pdfs. Connexions. 7 June 2005 . • Leon-Garcia, Alberto, and Alberto Leon-Garcia. Probability, Statistics, and Random Processes for Electrical Engineering. Upper Saddle River, NJ: Pearson/Prentice Hall, 2008. • Mood, Alexander McFarlane, Franklin A. Graybill, and Duane C. Boes. Introduction to the Theory of Statistics. New York: McGraw-Hill, 1973.

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