Math 362, Problem set 1

Transcription

1 Math 6, roblem set Due //. (4..8) Determine the mean variance of the mean X of a rom sample of size 9 from a distribution having pdf f(x) = 4x, < x <, zero elsewhere. We have that: Using the pdf, so E[ X] = E[ n Xi ] = n ne[x i] = E[X i ] var( X) = var( X i ) n = var(x i). n E[X] = E[X ] = var(x i ) = ( 4 5 4x 4 dx = x 5 dx =, var( X) = 675. ) = 75,. (4..5) Let S = n (Xi X) d denote the sample variance of a rom sample from a distribution with variance σ >. Since E[S ] = σ, why isn t E[S] = σ? Note: There is a hint in the book that gives it away, but maybe think about it for a second before looking there. This follows from the fact that f(x) = x is strictly convex (that is f (x) = >, from Jensen s inequality: σ = E[S ] > (E[S] ),

2 so E[S] < σ. Note: some of you also (cleverly) noticed this follows from the fact that var(s) >.. (5..4) Let X,..., X n be a rom sample from the Γ(, θ) distribution, where θ is unknown. (Recall, look back in chapter if you forget the specifics of the Γ distribution). Let Y = n i= X i. (a) Find the distribution of Y determine c so that cy is anunbiased estimator of θ. (b) If n = 5 show that (9.59 < Yθ ) < 4. =.95 (c) Using (b), show that if y is the value of Y once the sample is drawn then the interval ( ) y 4., y 9.59 is a 95% confidence interval for Θ. (d) Suppose the sample results in the values, Based on these data, obtain the point estimate of θ as described in art (a) the computed 95% confidence interval in art (c). What does the confidence interval mean? For (a), Y Γ(n, θ) by the additive property of the Gamma distribution. Thus E[Y ] = nθ we should take c = n. For (b), we will simply set up the integral. (9.59 < Yθ ) < 4. = (4.795θ < Y < 7.θ) = = 7.θ 4.795θ !θ x9 e x/θ dx u 9 9! e u du = Note: The important part here is that this does not depend on θ.

3 For (c), note that: (9.59 < Yθ ) < 4. = ( y 4. < θ < y 9.59 ) =.95 so the given interval is a 95% confidence interval. For (d), θ y n =.8654, we have a confidence interval of (6.685,.84), meaning that the true value of θ has a 95% chance of ling between these values. 4. (5..5) Suppose the number of customers X that enter a store between the hours of 9AM AM follows a oisson distribution with parameter θ. Suppose a rom sample of the number of customers for days results in the values Based on these data obtain an unbiased point estimate of θ. Explain the meaning of this estimate in terms of the number of customers. Since E[X] = θ for a oisson θ rom variable, we have a point estimate of 9.5 customers per day. (Of course while it is impossible to actually obtain this on any given day, but it s a fine estimate of the average - it s even perfectly fine if it is the true mean of the distribution. 5. (5..) Obtain the probability that an observationis a potential outlier for the following distributions (a) The underlying distribution is normal (b) The underlying distribution is logistic, in other words it has pdf: f(x) = e x ( + e x ), < x < (c) The underlying distribution is Laplace, the pdf given by: f(x) = e x, < x <. For (a), we find (from the back of the book) that Q =.675 Q =.675.

4 Thus h =.5, UF =.7 LF =.7. We have (Z > UF ) =.46697, so (LF < Z < UF ) =.9966, so the probability of a potential outlier is.7 For (b), we have that Q e x ( + e x ) dx = 4, similar for Q. We thus get Q = ln() Q = ln(). Thus h = ln(), UF = 4 ln() LF = 4 ln(). We have then, that the (LF < X < UF ) = 4 ln() 4 ln() e x 4 ( + e x dx = ) 4, so the probability of a potential outlier is For (c) we have that Q e x = 4. Solving, we find Q = ln() likewise Q = ln(). Thus h = ln(), we have UF = 4 ln() LF = 4 ln(). Since 4 ln() 4 ln() e x dx = 5 6. Thus the probability of a potential outlier is 6 = (5..5) Let Y < Y < Y < Y 4 be the order statistics of a rom sample of size 4 from the distribution having pdf f(x) = e x, < x <, zero elsewhere. Find ( Y 4 ). We have that F (x) = x e t dx = e t. Using the formula we derived in class: (Y 4 ) = 4( e t ) e t dt = ( e ) (5..) Let Y < Y < Y be the order statistics of a rom sample of size from a distribution having the pdf f(x) = x, < x <, zero elsewhere. Show that Z = Y /Y, Z = Y /Y Z = Y are mutually independent. 4

5 We have that Y = Z, Y = Z Z Y = Z Z Z. Thus, z z z z z z J = z z = z z. We have f Z,Z,Z (z, z, z ) = f Y,Y,Y (z z z, z z, z )z z =!(z z z )(z z )(z )(z z) = (z )(4z )(6z 5 ), where < z i <. This is the product of the marginal pdfs of Z, Z Z which shows that the variables are mutually independent. 8. (5..) Let X, X,..., X n be a rom sample. A measure of spread is Gini s mean difference n j ( ) n G = X i X j / j= i= (a) If n =, find a,..., a so that G = i= a iy i, where Y, Y,..., Y are the order statistics of the sample. (b) Show that E[G] = σ/ π if the sample arises from the normal distribution N(µ, σ ). Hint: This looks worse than it is: Think about the fact that X i X j is either X i X j or X j X i depending on which is larger. The Y i appear in the sum G, as they are some X j. The key observation is that Y i Y j = Y i Y j if i > j Y i Y j = Y j Y i, if i < j. Thus Y i appears with a positive sign i times, with a negative sign n i times. In all, Y i has coefficient: (i ) (n i) a i = ( n. ) Since n = 9, we have that for i =,..., 9.. a i = i 45 For (b), note that E[G] = E[ X i X j ]. Since X i X j N(, σ ) we have that E[ X i X j ] = = x x (π)(σ ) exp( σ )dx σ e u du = σ. π π 5