Discrete Random Variables

Transcription

1 Discrete Random Variables COMP 245 STATISTICS Dr N A Heard Contents 1 Random Variables Introduction Cumulative Distribution Function Discrete Random Variables Simple Random Variables Discrete Random Variables Probability Mass Function Mean and Variance Expectation Variance, Standard Deviation and Skewness Comments Sums of Random Variables Discrete Distributions Bernoulli, Binomial and Geometric Distributions Poisson Distribution Discrete Uniform Distribution

2 1 Random Variables 1.1 Introduction Definition Suppose, as before, that for a random experiment we have identified a sample space S and a probability measure P(E) defined on (measurable) subsets E S. A random variable is a mapping from the sample space to the real numbers. So if X is a random variable, X : S R. Each element of the sample space s S is assigned by X a (not necessarily unique) numerical value X(s). If we denote the unknown outcome of the random experiment as s, then the corresponding unknown outcome of the random variable X(s ) will be generically referred to as X. The probability measure P already defined on S induces a probability distribution on the random variable X in R: For each x R, let S x S be the set containing just those elements of S which are mapped by X to numbers no greater than x. That is, let S x be the inverse image of (, x] under the function X. Then noting the equaivalence we see that X(s ) x s S x. P X (X x) P(S x ). The image of S under X is called the range of the random variable: range(x) X(S) = {x R s S s.t. X(s) = x} So as S contains all the possible outcomes of the experiment, range(x) contains all the possible outcomes for the random variable X. Example Let our random experiment be tossing a fair coin, with sample space {H, T} and probability measure P({H}) = P({T}) = 1 2. We can define a random variable X : {H, T} R taking values, say, X(T) = 0, X(H) = 1. In this case, what does S x look like for each x R? if x < 0; S x = {T} if 0 x < 1; {H, T} if x 1. This defines a range of probabilities P X on the continuum R P( ) = 0 if x < 0; P X (X x) = P(S x ) = P({T}) = 1 2 if 0 x < 1; P({H, T}) = 1 if x 1. 2

3 1 P X (X x) x 1.2 Cumulative Distribution Function cdf The cumulative distribution function (cdf) of a random variable X, written F X (x) (or just F(x)) is the probability that X takes value less than or equal to x. F X (x) = P X (X x) For any random variable X, F X is right-continuous, meaning if a decreasing sequence of real numbers x 1, x 2,... x, then F X (x 1 ), F X (x 2 ),... F X (x). cdf Properties For a given function F X (x), to check this is a valid cdf, we need to make sure the following conditions hold F X (x) 1, x R; 2. Monotonicity: x 1, x 2 R, x 1 < x 2 F X (x 1 ) F X (x 2 ); 3. F X ( ) = 0, F X ( ) = 1. For finite intervals (a, b] R, it is easy to check that P X (a < X b) = F X (b) F X (a). Unless there is any ambiguity, we generally suppress the subscript of P X ( ) in our notation and just write P( ) for the probability measure for the random variable. That is, we forget about the underlying sample space and just think about the random variable and its probabilities. Often, it will be most convenient to work this way and consider the random variable directly from the very start, with the range of X being our sample space. 3

4 2 Discrete Random Variables 2.1 Simple Random Variables We say a random variable is simple if it can take only a finite number of possible values. That is, X is simple range(x) is finite. Suppose X is simple, and can take one of m values X = {x 1, x 2,..., x m } ordered so that x 1 < x 2 <... < x m. Each sample space element s S is mapped by X to one of these m values. Then in this case, we can partition the sample space S into m disjoint subsets {E 1, E 2,..., E m } so that s E i X(s) = x i, i = 1, 2,..., m. We can then write down the probability of the random variable X taking the particular value x i as It is also easy to check that where we can take x 0 =. P X (X = x i ) P(E i ). P X (X = x i ) = F X (x i ) F X (x i 1 ), Examples of Simple Random Variables Consider once again the experiment of rolling a single die. Then S = {,,,,, } and for any s S (e.g. s = ) we have P({s}) = 1 6. An obvious random variable we could define on S would be X : S R, s.t. X( ) = 1, X( ) = 2,. X( ) = 6. Then e.g. P X (1 < X 5) = P({,,, }) = 4 6 = 2 3 and P X (X {2, 4, 6}) = P({,, }) = 1 2. Alternatively, we could define a random variable Y : S R, s.t. Y( ) = Y( ) = Y( ) = 0, Y( ) = Y( ) = Y( ) = 1. Then clearly P Y (Y = 0) = P({,, }) = 1 2 and P Y(Y = 1) = P({,, }) =

5 Comments Note that under either random variable X or Y, we still got the same probability of getting an even number of spots on the die. Indeed, this would be the case for any random variable we may care to define. A random variable is simply a numeric relabelling of our underlying sample space, and all probabilities are derived from the associated underlying probability measure. 2.2 Discrete Random Variables A simple random variable is a special case of a discrete random variable. We say a random variable is discrete if it can take only a countable number of possible values. That is, X is discrete range(x) is countable. Suppose X is discrete, and can take one of the countable set of values X = {x 1, x 2,...} ordered so that x 1 < x 2 <.... Each sample space element s S is mapped by X to one of these values. Then in this case, we can partition the sample space S into a countable collection of disjoint subsets {E 1, E 2,...} s.t. s E i X(s) = x i, i = 1, 2,.... As with simple random variables we can write down the probability of the discrete random variable X taking the particular value x i as P X (X = x i ) P(E i ). We again have P X (X = x i ) = F X (x i ) F X (x i 1 ), For a discrete random variable X, F X is a monotonic increasing step function with jumps only at points in X. Example: Poisson(5) cdf F(x) x 5

6 2.3 Probability Mass Function For a discrete random variable X and x R, we define the probability mass function (pmf), p X (x) (or just p(x)) as p X (x) = P X (X = x) pmf Properties To check we have a valid pmf, we need to make sure the following conditions hold. If X can take values X = {x 1, x 2,...} then we must have: 1. 0 p X (x) 1, x R; 2. p X (x) = 1. x X Example: Poisson(5) pmf p(x) x Knowing either the pmf or cdf of a discrete random variable characterises its probability distribution. That is, from the pmf we can derive the cdf, and vice versa. p(x i ) = F(x i ) F(x i 1 ), F(x i ) = i p(x j ). j=1 6

7 Links with Statistics We can now see the first links between the numerical summaries and graphical displays we saw in earlier lectures and probability theory: We can often think of a set of data (x 1, x 2,..., x n ) as n realisations of a random variable X defined on an underlying population for the data. Recall the normalised frequency counts we considered for a set of data, known as the empirical probability mass function. This can be seen as an empirical estimate for the pmf of their underlying population. Also recall the empirical cumulative distribution function. This too is an empirical estimate, but for the cdf of the underlying population. 7

8 3 Mean and Variance 3.1 Expectation E(X) For a discrete random variable X we define the expectation of X, E X (X) = xp X (x). x E X (X) (often just written E(X) or even µ X ) is also referred to as the mean of X. It gives a weighted average of the possible values of the random variable X, with the weights given by the probabilities of each outcome. Examples 1. If X is a r.v. taking the integer value scored on a single roll of a fair die, then E(X) = 6 xp(x) x=1 = = 21 6 = If now X is a score from a student answering a single multiple choice question with four options, with 3 marks awarded for a correct answer, -1 for a wrong answer and 0 for no answer, what is the expected value if they answer at random? E(X) = 3.P(Correct) + ( 1).P(Incorrect) = = 0. E{g(X)} More generally, for a function of interest g : R R of the random variable X, first notice that the composition g(x), g(x)(s) = (g X)(s) is also a random variable. It follows that E X {g(x)} = g(x)p X (x) (1) x Linearity of Expectation Consider the linear function g(x) = ax + b for constants a, b R. We can see from (1) that E X (ax + b) = (ax + b)p X (x) x = a x xp X (x) + b x and since x xp X (x) = E(X) and x p X (x) = 1 we have p X (x) 8

9 E(aX + b) = ae(x) + b, a, b R. It is equally easy to check that for g, h : R R, we have E{g(X) + h(x)} = E{g(X)} + E{h(X)}. 3.2 Variance, Standard Deviation and Skewness Var(X) Consider another special case of g(x), namely g(x) = {X E(X)} 2. The expectation of this function wrt P X gives a measure of dispersion or variability of the random variable X around its mean, called the variance and denoted Var X (X) (or sometimes σ 2 X ): Var X (X) = E X [{X E X (X)} 2 ]. We can expand the expression {X E(X)} 2 and exploit the linearity of expectation to get an alternative formula for the variance. and hence {X E(X)} 2 = X 2 2E(X)X + {E(X)} 2 Var(X) = E[X 2 {2E(X)}X + {E(X)} 2 ] = E(X 2 ) 2E(X)E(X) + {E(X)} 2 Var(X) = E(X 2 ) {E(X)} 2. Variance of a Linear Function of a Random Variable We saw earlier that for constants a, b R the linear combination ax + b had expectation ae(x) + b. What about the variance? It is easy to show that the corresponding result is Var(aX + b) = a 2 Var(X), a, b R. sd(x) The standard deviation of a random variable X, written sd X (X) (or sometimes σ X ), is the square root of the variance. sd X (X) = Var X (X). 9

10 Skewness The skewness (γ 1 ) of a discrete random variable X is given by γ 1 = E X[{X E X (X)} 3 ] sd X (X) 3. That is, if µ = E(X) and σ = sd(x), Examples γ 1 = E[(X µ)3 ] σ If X is a r.v. taking the integer value scored with a single roll of a fair die, then Var(X) = 6 x 2 p(x) x=1 = = If now X is a score from a student answering a single multiple choice question with four options, with 3 marks awarded for a correct answer, -1 for a wrong answer and 0 for no answer, what is the standard deviation if they answer at random? 3.3 Comments E(X 2 ) = 3 2.P(Correct) + ( 1) 2.P(Incorrect) = = 3 sd(x) = = 3. Links with Statistics We have met three important quantities for a random variable, defined through expectation - the mean µ, the variance σ 2 and the standard deviation σ. Again we can see a duality with the corresponding numerical summaries for data which we met - the sample mean x, the sample variance s 2 and the sample standard deviation s. The duality is this: If we were to consider the data sample as the population and draw a random member from that sample as a random variable, this r.v. would have cdf F n (x), the empirical cdf. The mean of the r.v. µ = x, variance σ 2 = s 2 and standard deviation σ = s. 3.4 Sums of Random Variables Expectation of Sums of Random Variables Let X 1, X 2,..., X n be n random variables, perhaps with different distributions and not necessarily independent. Let S n = n X i be the sum of those variables, and S n n i=1 Then the mean of S n is given by be their average. 10

11 E(S n ) = n i=1 E(X i ), E ( ) Sn n = n i=1 E(X i). n Variance of Sums of Random Variables However, for the variance of S n, only if X 1, X 2,..., X n are independent do we have Var(S n ) = n i=1 Var(X i ), Var ( ) Sn = n i=1 Var(X i) n n 2. So if X 1, X 2,..., X n are independent and identically distributed with E(X i ) = µ X and Var(X i ) = σx 2 we get E ( ) ( ) Sn Sn = µ X, Var = σ2 X n n n. 11

12 4 Discrete Distributions 4.1 Bernoulli, Binomial and Geometric Distributions Bernoulli(p) Consider an experiment with only two possible outcomes, encoded as a random variable X taking value 1, with probability p, or 0, with probability (1 p), accordingly. (Ex.: Tossing a coin, X = 1 for a head, X = 0 for tails, p = 1 2.) Then we say X Bernoulli(p) and note the pmf to be p(x) = p x (1 p) 1 x, x = 0, 1. Using the formulae for mean and variance, it follows that Example: Bernoulli(¼) pmf µ = p, σ 2 = p(1 p). Binomial(n, p) Consider n identical, independent Bernoulli(p) trials X 1,..., X n. Let X = n i=1 X i be the total number of 1s observed in the n trials. (Ex.: Tossing a coin n times, X is the number of heads obtained, p = 1 2.) Then X is a random variable taking values in {0, 1, 2,..., n}, and we say X Binomial(n, p). From the Binomial Theorem we find the pmf to be p(x) = ( n x )px (1 p) n x, x = 0, 1, 2,..., n. ( ) n x n! To calculate the Binomial pmf we recall that = x x!(n x)! and x! = i=1 i. (Note 0! = 1.) It can be shown, either directly from the pmf or from the results for sums of random variables, that the mean and variance are Similarly, the skewness is given by µ = np, σ 2 = np(1 p). γ 1 = 1 2p np(1 p). 12

13 Example: Binomial(20,¼) pmf Example Suppose that 10 users are authorised to use a particular computer system, and that the system collapses if 7 or more users attempt to log on simultaneously. Suppose that each user has the same probability p = 0.2 of wishing to log on in each hour. What is the probability that the system will crash in a given hour? Solution: The probability that exactly x users will want to log on in any hour is given by Binomial(n, p) = Binomial(10, 0.2). Hence the probability of 7 or more users wishing to log on in any hour is Two more examples p(7) + p(8) + p(9) + p(10) ( ) ( ) = = A manufacturing plant produces chips with a defect rate of 10%. The quality control procedure consists of checking samples of size 50. Then the distribution of the number of defectives is expected to be Binomial(50,0.1). When transmitting binary digits through a communication channel, the number of digits received correctly out of n transmitted digits, can be modelled by a Binomial(n, p), where p is the probability that a digit is transmitted incorrectly. Note: Recall the independence condition necessary for these models to be reasonable. Geometric(p) Consider a potentially infinite sequence of independent Bernoulli(p) random variables X 1, X 2,.... Suppose we define a quantity X by X = min{i X i = 1} 13

14 to be the index of the first Bernoulli trial to result in a 1. (Ex.: Tossing a coin, X is the number of tosses until the first head is obtained, p = 1 2.) Then X is a random variable taking values in Z + = {1, 2,...}, and we say X Geometric(p). Clearly the pmf is given by p(x) = p(1 p) x 1, x = 1, 2,.... The mean and variance are µ = 1 p, σ2 = 1 p p 2. The skewness is given by γ 1 = 2 p 1 p, and so is always positive. Example: Geometric(¼) pmf Alternative Formulation If X Geometric(p), let us consider Y = X 1. Then Y is a random variable taking values in N = {0, 1, 2,...}, and corresponds to the number of independent Bernoulli(p) trials before we obtain our first 1. (Some texts refer to this as the Geometric distribution.) Note we have pmf p Y (y) = p(1 p) y, y = 0, 1, 2,..., and the mean becomes µ Y = 1 p p. while the variance and skewness are unaffected by the shift. 14

15 Example Suppose people have problems logging onto a particular website once every 5 attempts, on average. 1. Assuming the attempts are independent, what is the probability that an individual will not succeed until the 4 th? p = 4 5 = 0.8. p(4) = (1 p)3 p = = On average, how many trials must one make until succeeding? Mean= 1 p = 5 4 = What s the probability that the first successful attempt is the 7 th or later? p(7) + p(8) + p(9) +... = p(1 p)6 1 (1 p) = (1 p)6 = Example (contd. from Binomial) Again suppose that 10 users are authorised to use a particular computer system, and that the system collapses if 7 or more users attempt to log on simultaneously. Suppose that each user has the same probability p = 0.2 of wishing to log on in each hour. Using the Binomial distribution we found the probability that the system will crash in any given hour to be Using the Geometric distribution formulae, we are able to answer questions such as: On average, after how many hours will the system crash? Mean = 1 p = = 1163 hours. Example: Mad(?) Dictators and Birth Control A dictator, keen to maximise the ratio of males to females in his country (so he could build up his all male army) ordered that each couple should keep having children until a boy was born and then stop. Calculate the number expected number of boys that a couple will have, and the expected number of girls, given that P(boy)=½. Assume for simplicity that each couple can have arbitrarily many children (although this is not necessary to get the following results). Then since each couple stops when 1 boy is born, the expected number of boys per couple is 1. On the other hand, if Y is the number of girls given birth to by a couple, Y clearly follows the alternative formulation for the Geometric(½) distribution. So the expected number of girls for a couple is = 1. 15

16 4.2 Poisson Distribution Poi(λ) Let X be a random variable on N = {0, 1, 2,...} with pmf for some λ > 0. p(x) = e λ λ x, x = 0, 1, 2,..., x! Then X is said to follow a Poisson distribution with rate parameter λ and we write X Poi(λ). Poisson random variables are concerned with the number of random events occurring per unit of time or space, when there is a constant underlying probability rate of events occurring across this unit. Examples the number of minor car crashes per day in the U.K.; the number of mistakes on each of my slides; the number of potholes in each mile of road; the number of jobs which arrive at a database server per hour; the number of particles emitted by a radioactive substance in a given time. An interesting property of the Poisson distribution is that it has equal mean and variance, namely µ = λ, σ 2 = λ. The skewness is given by γ 1 = 1 λ, so is always positive but decreasing as λ. Example: Poi(5) pmf (again) 16

17 Poisson Approximation to the Binomial Notice the similarity between the pmf plots we ve seen for Binomial(20,¼) and Poi(5). It can be shown that for Binomial(n, p), when p is small and n is large, this distribution can be well approximated by the Poisson distribution with rate parameter np, Poi(np). p in the above is not small, we would typically prefer p < 0.1 for the approximation to be useful. The usefulness of this approximation is in using probability tables; tabulating a single Poisson(λ) distribution encompasses an infinite number of possible corresponding Binomial distributions, Binomial(n, λ n ). Ex. A manufacturer produces VLSI chips, of which 1% are defective. Find the probability that in a box of 100 chips none are defective. We want p(0) from Binomial(100,0.01). Since n is large and p is small, we can approximate this distribution by Poi( ) Poi(1). Then p(0) e 1 λ 0 0! = Fitting a Poisson Distribution to Data Ex. The number of particles emitted by a radioactive substance which reached a Geiger counter was measured for 2608 time intervals, each of length 7.5 seconds. The (real) data are given in the table below: x n x Do these data correspond to 2608 independent observations of an identical Poisson random variable? The total number of particles, x xn x, is 10,094, and the total number of intervals observed, n = x n x, is 2608, so that the average number reaching the counter in an interval is = Since the mean of Poi(λ) is λ, we can try setting λ = 3.87 and see how well this fits the data. For example, considering the case x = 0, for a single experiment interval the probability of observing 0 particles would be p(0) = e ! = So over n = 2608 repetitions, our (Binomial) expectation of the number of 0 counts would be n p(0) = Similarly for x = 1, 2,..., we obtain the following table of expected values from the Poi(3.87) model: x O(n x ) E(n x ) (O=Observed, E=Expected). 17

18 The two sets of numbers appear sufficiently close to suggest the Poisson approximation is a good one. Later, when we come to look at hypothesis testing, we will see how to make such judgements quantitatively. 4.3 Discrete Uniform Distribution U({1, 2,..., n}) Let X be a random variable on {1, 2,..., n} with pmf p(x) = 1, x = 1, 2,..., n. n Then X is said to follow a discrete uniform distribution and we write X U({1, 2,..., n}). The mean and variance are and the skewness is clearly zero. µ = n + 1, σ 2 = n