Political Science Math Camp: Problem Set 2

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Ralph Sparks
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1 Political Science Math Camp: Problem Set 2 Due Thursday Aug at 9:00 am. Suppose the probability of observation O given hypothesis H is P (O H ) = 3, the probability of observation O 2 given H is P (O 2 H ) = 2 3, the probability of observation O given hypothesis H 2 is P (O H 2 ) = 2 3 and the probability of observation O 2 given H 2 is P (O 2 H 2 ) = 3. The prior probabilities of hypotheses H and H 2 are P (H ) = P (H 2 ) = 2. Research then yields observation O. Given this observation, what is the probability of H? That is, what is P (H O )? P r(o H ) = 3 P r(o 2 H ) = 2 3 P r(o H 2 ) = 2 3 P r(o 2 H 2 ) = 3 P r(h ) = 2 P r(h 2 ) = 2 P r(h O ) = P r(o H )P r(h ) P r(o ) ( 3 = )( 2 ) ( 3 )( 2 ) + ( 2 3 )( 2 ) = = The random variable X can take values 2, 0 and 2, with probabilities 6, 6 and 4 6 respectively. What is the expected value of X? What is the variance of X? E(X) = 2( 6 ) (4 6 ) = =

2 V ar(x) = ( 2 ) 2 ( 6 ) + (0 )2 6 + (2 )2 4 6 = ( 9 6 ) + ( 6 ) + (4 6 ) = 4 6 = The random variable Z is distributed according to the density function: 0 ifz < f(z) = h( z 2 ) if z 0 if < z where h is a parameter to be determined. (a) What is the value of h? h h( z 2 )dz = [ (z 3 z3 ) ] = h[( 3 ) ( + 3 )] = h( ) = h( 4 3 ) = h = 3 4 (b) Find the cumulative distribution function F (z) for any z between and F (z) = = z z f(z)dz 3 4 ( x2 )dx (c) What is the probability that z is between 0 and 2? 2

3 First Draw Second Draw n.a., 2, 3, 4, 5, 6 2 2, n.a. 2, 3 2, 4 2, 5 2, 6 3 3, 3, 2 n.a. 3, 4 3, 5 3, 6 4 4, 4, 2 4, 3 n.a. 4, 5 4, 6 5 5, 5, 2 5, 3 5, 4 n.a. 5, 6 6 6, 6, 2 6, 3 6, 4 6, 5 n.a. Table : In the table, 2 means that is the observed value of the first draw and 2 is the observed value of the second draw. P r(0 z 2 ) = ( z2 )dz = 3 [ (z ] z3 ) = 3 4 ( 2 24 ) 3 (0 0) 4 = A ticket is drawn at random from a box containing six tickets:, 2, 3, 4, 5, 6. Then a second ticket is drawn, without replacement of the first ticket. (a) What is the probability that the second ticket is 3? What is the probability that the second ticket is 3, given that the first ticket is 2? One way to come up with this answer is to plot out all the ways in which two draws of the tickets can come out (see Table ). Note that there are 6x5=30 ways of drawing one ticket and then another. In 5 of these 30 outcomes depicted in Table, the second draw is a 3. Thus, the probability is 5/30 or /6. Notice this is the same as the probability of drawing a 3 on the first draw. Another way to arrive at this answer is to reason as follows: to get a 3 on the second draw you have to NOT get a 3 on the first. The probability of not getting a 3 is 5/6 on the first draw. Then the box is reduced to {X, X, X, X, 3} and the probability of drawing a 3 on the second is /5. These probabilities dependent, so we need to use conditional probabilities (see FPP Ch. 3 and 4). Thus, P(Second = 3) = P(First 3) P(Second = 3 P(First 3) = = 6 As for the probability that the second ticket is 3, given that the first ticket is 2, look at the second row of Table. This row shows all the ways for the second draw to come out if the first ticket is a 2. Now, the box for the second draw is {, 3, 4, 5, 6}. Since each number is equally likely to be drawn, the probability that the draw will be a 3 is then /5. If this material is unfamiliar, review Chapter 3 and 4 of FPP. (b) Is the unconditional probability the same as the conditional probability? Is the value of the second ticket dependent or independent of the value of the first ticket? 3

4 Unconditional probability is just the probability that an event occurs without reference to any other event. A conditional probability is the probability that an event occurs given that something else has transpired. For example, the first question in part (a) refers to an unconditional probability, while the second question refers to a conditional probability. If the draws are independent, unconditional and conditional probabilities are the same; here, they are different because we are drawing without replacement, so the draws are dependent. Knowing the value of the first ticket tells you something about the chances of pulling a certain number on the second draw. (c) What is the probability that the first ticket is 2 and the second ticket is 3? What is the probability that the first or the second ticket is 3? Explain your answers. If this material is unfamiliar, read Chapters 3 and 4 of Freedman, Pisani, and Purves (2007; hereafter FPP). Table shows that the first ticket is 2 and the second ticket is 3 only once out of 30 outcomes; since each of the 30 outcomes is equally likely, the probability is 30. Alternately, use the formula for conditional probability: P(First = 2 & Second = 3) = P(First = 2) P(Second = 3 First = 2). () Now, P(First = 2) = 6. (2) Next, if the first ticket drawn is a 2, the box becomes {,3,4,5,6} so P(Second = 3 First = 2) = 5 (3) Thus, P(First = 2 & Second = 3) = 6 5 = 30. (4) As for the probability that the first OR second ticket is 3, Table shows that there are five ways for the first draw to be a 3 (third row of Table ), given that we are drawing twice without replacement, and five ways for the second draw to be a 3 (third column of Table ). Thus, the probability is 0 30 or 3. Notice that these are mutually exclusive events: if the first ticket is 3, the second ticket cannot be (we are drawing without replacement), and if the second ticket is 3, the first ticket was not. Thus, the probability is the sum of the individual events (FPP Ch. 4.2). The probability that the first ticket is 3 is 6, and the probability that the second ticket is 3 is also 6. Thus, the probability that the first or second ticket is 3 is = Suppose we have a box with two tickets, one a 0 and the other a. We draw at random with replacement from this box n times. (a) What is the standard deviation of the box? (See FPP Ch and Ch. 7.4). Use the short cut described in FPP Ch Here the big number is and the small number is 0. The fraction of tickets with is 2, and so is the fraction with 0. So the standard deviation is ( 0) 2 2 = 2. (Note that X Y means the product of X and Y ). Slightly more formally, the box has a binomial distribution with mean p. So the variance is p( p) = 2 ( 2 ) = 4 and the s.d. is thus 2 2 = 2. 4

5 Finally, you could also do this the long way: the s.d. is the square root of the mean squared deviations from the average. Here, that s 2 [( 2 )2 + (0 2 )2 = 2 2 = 2. (5) The long way is not so long here, because there are only two tickets in the box. (b) What is the variance of a single ticket drawn from the box? (Note: this question asks you about the variance of a random variable.) The variance of a single draw is the variance of the box (see e.g. Freedman and Lane). Thus, the variance (the square of the standard deviation) is 4. (c) Consider the average of n tickets drawn from the box. Is this a random variable? What is its variance? What is its standard deviation? How does the standard deviation relate to the standard error of the mean? Drawing randomly with replacement from the box is a chance process. Since the realization of each draw is a random variable, the average of these realizations is also a random variable. In part (b), we noted that the variance of a single draw is the variance of the box. Denote the realization of a single draw as Y i. The variance of this random variable is Var(Y i ). Finally, the average of n tickets drawn from the box is Y = n n i= Y i. Thus, Var(Y ) = Var( n n Y i ) i= = n n 2 Var(Y i ) i= = nvar(y i) n 2 = Var(Y i). (6) n In the second line, we can distribute the variance inside the sum because the Y i s are independent (we are drawing each Y i with replacement). (Note also that the constant n becomes when n 2 taken outside of the variance). And for the third line, the distribution of each draw Y i is the same: we are drawing with replacement, so the contents of the box (i.e., its distribution) don t change from draw to draw. Thus, Var(Y i ) is the same for each i =,..., n. So, n i= Var(Y i) = nvar(y i ). Cancelling the ns gives the result. Finally, since Var(Y i ) is 4 (from part b), Var(Y ) is 4n, and the standard deviation is the square root: 2. Now comes a final idea. The sampling variance of the mean that is, Var(Y ) is a n theoretical quantity: it is a measure of the spread of the Y s across hypothetical replications of this process of drawing n tickets at random from the box. The distribution of all of the means across these hypothetical replications is called the sampling distribution of the mean. Var(Y ) is the variance of that distribution, and Var(Y ) is the standard deviation. However, the standard deviation of the random variable Y is called the standard error of the mean. 6. Now suppose we have a box with three tickets, numbered, 2, and 3. We draw two tickets at random without replacement. (a) Create a list or table showing all the ways that the two draws can come out. For each of the possible outcomes, note the average of the two tickets. Show that the average of these averages equals the average of the box. 5

6 The box is [, 2, 3]. The average of the box is 2. Let X i be the first draw and Y i be the second draw. There are i =,..., 6 possible orders in which these the tickets in the box can be drawn. i {X i, Y i } X i +Y i 2 ) {,2}.5 2 {,3} 2 3 {2,}.5 4 {2,3} {3,2} {3,} 2 Average: 2 Thus, the average of all possible draws equals the average of the box. (b) For all of the possible outcomes, note the value of the first ticket drawn. What is the average of these values? The average of all possible values for the first ticket is 6 6 i= (X i) = 2 (c) For all the possible outcomes, note the value of the second ticket drawn. What is the average of these values? The average of all possible values for the second ticket drawn is 6 6 i= (Y i) = 2 (d) Are each of these possible outcomes equally likely? Why or why not? Before the sampling begins, each of these 6 possible outcomes is equally likely. (e) Does the expected value of the first draw equal the expected value of the second draw? Why or why not? Yes, the expected value of the first draw is 2, and so is the expected value of the second draw by parts (b)-(d). In more detail, the average of the first ticket drawn, across the six possible outcomes of two draws, is 2, and so is the average of the second ticket drawn, by parts (b-c). By (d), each of these six outcomes is equally likely. The lesson: we are drawing without replacement, but the expected value of every draw is the same. (If we knew the value of the first draw, then the expected value of the second draw would be affected: that s the difference between conditional and unconditional expectations). Notice that we invoke this result when we write e.g. E(Y ) = E( n n i= Y i) = n n i= E(Y i) = ne(y i ) n = E(Y i ): in the penultimate step, the expected value of every draw is the same, so we can sum these constants. The expected value of the mean is the expected value of a single draw, which is the average of the box. α = x + y = x y 6

Example. If 4 tickets are drawn with replacement from ,

Example. If 4 tickets are drawn with replacement from 1 2 2 4 6, what are the chances that we observe exactly two 2 s? Exactly two 2 s in a sequence of four draws can occur in many ways. For example, (