Continuous Random Variables

Transcription

1 1 Continuous Random Variables Example 1 Roll a fair die. Denote by X the random variable taking the value shown by the die, X {1, 2, 3, 4, 5, 6}. Obviously the probability mass function is given by (since all outcomes are equally likely) k P(X = k) 1/6 1/6 1/6 1/6 1/6 1/6 Consider, for instance, the event X 4. Since all the events X = k are pairwise disjoint the probability P (X 4) can be obtained by adding probabilities, and we clearly obtain P (X 4) = 4 1/6 = 2/3. We certainly can consider such probabilities as well for non integer values, e.g., P(X 4.2) = 4 as well. In more general terms F X (t) = P(X t) defines a function on R, the so called cumulative distribution function. It can be obtained from the pmf, and written down in a piecewise way F X (t) = 0 if t < 1 F X (t) = n 6 if n t < n + 1, n = 1, 2,..., 5 F X (t) = 1 if t 6 Cumulative distribution function for rolling a fair die: F (t) X 1 5/6 2/3 1/2 1/3 1/ t In this example (and in general for discrete random variables) the cumulative distribution function is discontinuous. The jumps/discontinuities occur at the values where the pmf is defined and the hight of the jump is the corresponding value of the pmf. The previous example motivates the following definition

2 2 Definition 1 The cumulative distribution function (cdf) F X of a random variable X is defined by the expression F X (t) = P(X t) for any real value t. Lemma 1 For any pair of values s < t the cumulative distribution function obeys F X (t) F X (s) = P(s < X t). Proof: Observe that the events X s and s < X t are pairwise disjoint, so that (using definition 1) F X (s) + P(s < X t) = P(X s) + P(s < X t) = P(X s or s < X t) = P(X t) = F X (t) The statement of the lemma follows by rearrangement. Remark: Definition 1 and lemma 1 have a couple of consequences: 0 F X (t) 1 since probabilities take values between 0 and 1. The cdf F X is an increasing function. If s < t then lemma 1 and the non negativity of probabilities tell us that F X (t) F X (s) 0, i.e. F X (s) F X (t) (which is the definition of an increasing function). If X is a discreet random variable taking values x 1, x 2,... then F X is a discontinuous function. Discontinuities (jumps) occur at t = x k with hight P(X = x k ). A formal proof of this assertion (which requires background knowledge in analysis) uses lemma 1 and one sided limits in s and t. Example 2 You roll two dice. The largest number shown is a random variable which we denote by X. Compute probabilities by ordered sampling with replacement. The size of the sample space is 6 2. If N is any integer between 1 and 6 (N = 1, 2, 3, 4, 5, 6) the event X N has size

3 3 N 2 (as there are N choices for each die). Hence F X (N) = P(X N) = (N/6) 2 and the cumulative distribution function (for any real t) reads F X (t) = 0 if t < 1 F X (t) = N if N t < N + 1, N = 1, 2,..., 5 F X (t) = 1 if t 6 The cdf has discontinuities (jumps) at x N = N for N = 1, 2,..., 6 of hight F X (N) F X (N 1) = (N/6) 2 ((N 1)/6) 2 = (2N 1)/6 2 = P (X = N) and hence the pmf in this case reads N P(X = N) 1/36 1/12 5/36 7/36 1/4 11/36 Hence there are cases where the cdf can be used to obtain the pmf. Let us explore how the setup can be adapted to cases where a random variable takes values from a whole range of real numbers (i.e. a variable which is not confined to a discrete set). Example 3 A bus runs every 10 minutes. You arrive at a bus stop. Let T denote the time you wait for the next bus. T is a random variable which can take any real value between 0 and 10. Obviously P(T 10) = 1, P(T x) = 0 if x is a negative value, and P(T x) = 1 if x is larger than 10. Hence we have partial information about the cdf. Assume further that the bus is equally likely to come at any time in the next 10 minutes. To be more precise: we mean that the probability for the bus to arrive between now and time x (with x < 10), P(T x), is proportional to x, i.e., P (T x) = x/10 (to comply with the aforementioned constraint P(T 10) = 1). Using definition 1 we summarise our findings for the cdf of the continuous random variable T

4 4 F T (x) = 0 if x < 0 x/10 if 0 x 10 1 if x > 10 Cumulative distribution function for bus waiting time: F (x) T 1 1/ x The example shows that the cdf F T function. of the continuous random variable T is a continuous In general Definition 2 A random variable T is called a continuous random variable if the cdf F T (x) is a continuous function Lemma 1 tells us the relation between the cdf F T and the probability of events, F T (x) F T (y) = P(y < T x) for y < x. If the take the limit y x (i.e. y approaching x) then the left hand side, F T (x) F T (y), tends to zero (because F T is continuous) whereas the right hand side, P(y < T x), tends to P (T = x). Hence, for continuous random variable T the probability that the variable takes a value, say T = x, is zero (that makes perfect sense, e.g., the probability that a bus arrives precisely after 1 minute seconds is zero). Hence, no probability mass function can be defined. Nevertheless, using the fact that the cdf F T (x) has a derivative for almost all values of x (see e.g. example 3) we can define a so called probability density function instead. Definition 3 The probability density function (pdf) of a continuous random variable T is defined by the derivative of the cumulative distribution function: f T (x) = df T (x) dx.

5 5 Remark: Definition 3 has a couple of consequences following from properties of the cdf: The probability density function is non negative, f T (x) 0, since the cdf is an increasing function (i.e. its slope is non negative). Using lemma 1 and the fundamental theorem of calculus we have P(y < T x) = F T (x) F T (y) = y x f T (s)ds for y < x. Hence the probability that the random variable T takes values in the interval (y, x] is given by the size of the area under the graph of the probability density function (see, e.g., example 4). In particular f T (s)ds = 1 since the random variable T takes a real value so that P( < T < ) = 1. Example 4 Consider the setup of example 3. According to definition 3 the probability density function of the bus waiting time reads: f T (x) = 0 if x < 0 1/10 if 0 x 10 0 if x > 10 Probability density function for bus waiting time: f (x) T 1/10 1/ x The shaded area illustrates the probability computed below. The density function is constant. We say that T has a uniform distribution.

6 6 The probability that one waits for the bus more than 7 minutes but less than 11 minutes can be evaluated as P(7 < T 11) = 11 7 f T (x)dx = 10 7 f T (x)dx = f T (x)dx = dx dx = With a given probability density function computing probabilities becomes integration.