II. Random Variables

Transcription

1 II. Random Variables Random variables operate in much the same way as the outcomes or events in some arbitrary sample space the distinction is that random variables are simply outcomes that are represented numerically. A ith t d t d ibl itlltt As with events, we denote a random variable as a capital letter, such as, Y, or Z. Realizations of a random variable are denoted using lower case. For instance, x may be used to represent a possible value of the variable.

2 Variable Types Random Variables Discrete/Categorical (e.g., ethnic group, gender) Continuous (e.g., age, height, weight)

3 Variables Map Outcomes to Numbers Experimental Outcomes Possible Numeric Values of a Random Variable O 1 x 1 O x O x 3 3 O 4 x 4

4 Example II.A Consider an experiment where we flip a coin: the sample space is S = {H, T}. We can define a random variable such that = 1 if the flip turns up heads and = 0 if the flip is tails. Example II.B Consider an experiment where we sample an adult and ask for his/her age in years: we might view the sample space as something like S = {18, 19, 0,,10}. Suppose we let Y = the person s age in years. Then in this case the mapping of outcomes in S to numeric values is straightforward.

5 Categorical Random Variables Possible values of a categorical random variable have probabilities biliti determined by a probability mass function, or pmf. Suppose that x 1,x,,x N are the N possible outcomes of a random variable. The pmf for, denoted by f(x) or P( = x), must have the following properties: i. 0 P( xi ) 1, for i 1,, N, and N ii. P( ) 1. i1 x i

6 Example II.C Let be the sum of two fair six-sided dice. Then the pmf for is given by x i P( = x i ) P( <= x i ) 1/36 1/36 3 /36 3/36 4 3/36 6/36 5 4/36 10/36 6 5/36 15/36 7 6/36 1/36 8 5/36 6/36 9 4/36 30/ /36 33/36 11 /36 35/36 1 1/36 1

7 Example II.C (cont d) The third column in the previous table contains the cumulative distribution function (cdf), or P( x i ). Note that the pmf determines the cdf, and vice versa. The pmf and cdf are plotted below: CDF PDF P( = x) P 0.0 ( <= x) x x

8 Continuous Random Variables As opposed to a probability mass function, a continuous random variable has a probability density function, or pdf. Suppose a continuous random variable can take any value between l and u. We denote the pdf of by f(x), for l < x < u. However, note that f(x) P( = x). Why must this be so? How then do we interpret the pdf for a continuous random variable?

9 Interpreting the Probability Density Function It doesn t make sense with a continuous variable to talk about P( = x). However, we can use the pdf f(x), for l < x < u, to compute the probability that we observe a value for that is within an interval, for example P(a b). Note: the probability P(a b) is given by the area under the curve f(x) between the points a and b, where a b and a and b are possible values of. Question: how does one find the area under a continuous curve between two points?

10 Properties of the Probability Density Function Given a pdf f(x), ) we can compute the probability bilit over an interval by P ( a b ) b a f ( x ) dx. This probability has two interpretations: 1. It represents the probability that a randomly selected individual from the underlying gpopulation p will have a value of that falls within the interval (a, b).. It represents the proportion of individuals in the underlying population who have values for that fall within the interval (a, b).

11 Properties of the Probability Density Function Hence, a pdf dff( f(x) ) must thave the following properties: i. f(x) 0, for l x u. u l ii. f ( x) dx 1. In other words, a pdf f(x) must be positive over its domain (l, u), and the area under the curve over that domain must be equal to 1.

12 Example II.D Stat 3000 Statistics for Scientists and Engineers Victims of a particular type of cancer have post-diagnosis survival times that follow the pdf plotted on the following slide. If represents the survival time (in months) of a given cancer patient, then the pdf for is f ( x) 1 15 Verify that this is a legitimate pdf. e x /15 On the plot, diagram the area that represents the probability that a given cancer patient survives no longer than 10 months. Diagram the area that represents the proportion who survive between 1 and 18 months. Diagram the area representing the proportion who survive longer than 18 months., x 0.

13 Example II.D, cont d Probability density function for survival time of cancer patients. 0.0 PDF P.Df(x).F. of of Chip Surviv Lifeti val me Time -- f(x) Survival Time After Diagnosis (months) Lifetime of Chip -- x

14 Example II.D, cont d Compute the probabilities that you diagrammed on the plot of the pdf. Find and plot the cumulative distribution function. Use the cdf to compute the probabilities that you diagrammed on the plot of the pdf. What is the median survival time? What is the 75 th percentile of survival?

15 The Mean or Expectation of a Random Variable Based on Example II.C, suppose that t I propose the following game: you pay me $7.50, then roll the dice. I pay you the sum of the dice in dollars. From a purely economic standpoint, should you play? In order to answer that question you need to know something In order to answer that question, you need to know something about the mean of, or the average sum of the dice.

16 Definition of Expectation By definition, iti the average or expected tdvalue of a categorical random variable with pmf at a value x i given by P(= x i ) for i = 1,,N is computed as E( ) N i1 x P( i x i The expected value of a continuous random variable with pdf given by f(x), for l x u, is computed as ). E( ) u l xf ( x) dx. I ith i th t ti i ht d In either case, we can view the expectation as a weighted average.

17 Example II.E The expected value of the distribution in Example II.C is given by E ( ) (1/ 36) 3( / 36) 1(1/ 36) 7. Playing the game described on the previous slide probably wouldn t be your smartest move that is, what would be your average net gain (or loss)? In general, how do we interpret expectation? How would you describe expectation to the average person? (Hint: recall our discussion of long-term relative frequency.)

18 Example II.F What is the average survival time for the cancer patients in Example II.D? The expectation in this case is computed by x /15 ( ) ( ) /15 /15 x x x E xf x dx e dx xe 15e x /15 x /15 lim xe 15e (0 15) x. Hence, the average life expectancy after diagnosis is 15 months. How do the mean and median compare? Why are they different?

19 The Variance of a Random Variable The mean is used to describe the center of a distribution. The variance is used to describe the spread of a distribution about its mean. As shown in the plot below, two distributions with the same mean can have very different degrees of variability about the mean:

20 Interpreting the Variance The variance is actually all the average squared deviation of the possible values of a random variable about its mean. In some sense, you can think of variance probabilistically: (i) If you sample one observation at random from a distribution with high variance, then there is a relatively high probability that the observation will lie far away from the mean. (ii) If you sample an observation from a distribution with low variance, then there is a relatively small probability that the observation will lie far from the mean.

21 Definition of the Variance By definition, the variance of a categorical random variable y, g with pmf at a value x i given by P(= x i ) for i = 1,,N is computed as N N Th i f ti d i bl ith df i b. )] ( [ ) ( ) ( )] ( [ ) ( 1 1 E x P x x P E x Var N i i i N i i i The variance of a continuous random variable with pdf given by f(x), for l x u, is computed as )} ( { ) ( ) ( )} ( { ) ( E d f d f E V u u. )} ( { ) ( ) ( )} ( { ) ( E dx x f x dx x f E x Var l l Note that the last part of the equations given above result from the fact that Var() = E( ) {E()}. This is typically much easier to compute by hand.

22 Example II.G For the distribution in Example II.C, Var ( ) (1/ 36) 3 ( / 36) 1 (1/ 36) Hence, the standard deviation is.4.

23 Example II.H For the distribution of survival times in Example II.D, x /15 ( ) ( ) x E x f x dx e dx x lim x e x /15 15xe x /15 ()15 e x /15 x /15 x /15 x /15 x e 15xe ()15 e {0 0 ()15 } ()15 ()15 The variance then is Var() = E( ) {E()} = (15) 15 = 15 (months). Hence, the standard deviation is 15 months.. 0

24 Some Additional Properties of the Mean and Variance In practical research settings, we are almost always interested t in the distribution of sums of random variables. In particular, as we shall see within a few weeks, we need to know something about the mean and variance of the sample mean 1 n n i 1 where the values 1,, n represent a sample of observations from some population lti (e.g., we sample 0USU students td t and measure their individual heights). Question: if each subject in the sample has a mean E( i ) = μ, and a variance Var( i ) = σ, then what are the mean and variance of? i,

25 Mean and Variance of a Linear Function Let be a random variable such that E() ( ) = μ and Var() ( ) = σ. Suppose we re interested in a random variable Y defined as Y = a + b, where a and b are constants. In other words, Y is a linear function of. Then E(Y) = E(a + b) = E(a) + be() = a + bμ, and Var(Y) = Var(a + b) = Var(a) + b Var() = 0 + b σ = b σ.

26 Example II.I The average high temperature during the month of September in Logan, UT, is 67.6 F, with a standard deviation of 4. F. That is, if is a variable that represents the high temperature in Logan on a randomly selected September day, then E() = 67.6 and Var() = 4.. What is the average daily temperature in degrees Celsius? What is the standard deviation of daily temperature in degrees Celsius?

27 Mean and Variance of a Linear Combination of Independent Random Variables Let 1 be a random variable such that E( 1 ) = μ 1 and Var( 1 ) = σ 1, and be a random variable such that E( ) = μ and Var( ) = σ, where 1 and are independent. Suppose we re interested in a random variable Y defined as Y = c c, where c 1 and c are constants. In other words, Y is a linear combination of 1 and. Then E(Y) = E(c c ) = c 1 E( 1 ) + c E( ) = c 1 μ 1 + c μ, and Var(Y) = Var(c c ) = c 1 Var( 1 ) + c Var( ) = c 1 σ 1 + c σ. Note that the expectation of Y given above holds regardless of the dependence of 1 and. However, 1 and must be independent for the variance formula to hold.

28 Example II.J Consider the cancer patients whose survival follows the distribution in Example II.D. Suppose that we randomly sample individuals with this cancer, and measure their survival after diagnosis. What is the expected combined survival of these individuals, id in months? In years? What is the standard deviation of their combined survival, in months? In years?

29 The Mean and Variance of a Sample Mean It might sound odd to talk about the mean and variance of a mean, but recall how a sample mean (which is just a simple average) is computed: we randomly sample n subjects 1,, n from a population with an underlying expected value of μ and variance of σ. That is, E( i ) = μ and Var( i ) = σ, for i = 1,,n. The sample mean is defined as 1 n n i 1 In other words, the sample mean is just a linear combination of the independent random variables 1,, n. Hence, the sample mean is also random variable, and so ith has a distribution ib ti of fits own, with a mean and and a variance. i,

30 The Mean and Variance of a Sample Mean Note the distinction between (i) a sample mean and (ii) the expectation and variance of the underlying distribution (we can t emphasize this enough): Sample Mean Random Variable and Expectation and Variance Fixed Parameters (Constants)

31 The Mean and Variance of a Sample Mean Having sampled the n subjects 1 1,,,, n at random from a population with an underlying expected value of μ and variance of σ, we can assume that these subjects are independent. Then using the rules given four slides previous the expectation of the distribution of the sample mean is μ, and variance of is σ /n. Can you demonstrate why?

32 Example II.K Consider again the cancer patients whose survival follows the distribution in Example II.D. Suppose that we randomly select 0 individuals with this cancer, and compute the average survival time for the sample. What is the expected value of? What is the variance of? By what factor do we need to increase our sample size to divide the standard deviation of in half?