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# 5 - MIT OpenCourseWare http/ocw.mit.edu 14.30 Introduction...

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MIT OpenCourseWare http://ocw.mit.edu 14.30 Introduction to Statistical Methods in Economics Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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Problem Set #5 - Solutions 14.30 - Intro. to Statistical Methods in Economics Instructor: Konrad Menzel Due: Tuesday, March 31, 2009 Question One The convolution formula is a useful trick when we are interested in the sum or average of independent random variables. In the last problem set, we dealt with the random variable X, below. e-" forx > 0 f (4 = 0 forx<O Now, suppose that X = X1 = X2 = -. - = XI, are independent, identically distributed random variables. 1. Using the convolution formula, determine the PDF of & = (xl + X2). Hint: Defilze Zl = X1 and Z2 = X1 + X2 and then use the transformation method to get back to Y2 from Z2. Solution to 1: To use the convolution formula, we need the joint PDF of X1 and X2 and x2 as a function of y2 and xl. The function is x2 = 2y2 - xl. Also, since they are independent, we can just construct the joint PDF by multiplying the two marginals, fxl(xl) and fx2(x2). This gives a joint PDF of f (xl, x2) = e-(x1+x2) for xl > 0 and x2 > 0 and zero otherwise. The convolution formula adapted to this problem (taking into account the limits) is (where we take into account the change in variables formula for x2 as a function of xl and y2 using the Jacobian) 2. Compute its expectation: IE[Y2].
Solution to 2: To compute the expectation, we use the standard formula for continuous random variables: This integral must be computed using two iterations of integration by parts. Alternatively, we could have used the properties of expectations (no convolution necessary) to obtain 3. Using the convolution formula, determine the PDF for & = ;(XI + X2 + X3). Hint: Use the hint from part 1 to define Z3 = XI + X2 + X3 and perform a convolution with X3 and Z2 to transform the problem into Z2 and Z3.

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Solution to 3: Apply the same method to the results from part I, where we obtained fY2(y2).We write the joint PDF of XI, X2, and X3 as We'll take a slightly different approach for this problem, in order to help us solve part 5 later. Define Z2 = XI +X2 and Z3 = Xl +X2 +X3. Now, we can ignore any Jacobian transformation and directly apply the convolution formula. We can first solve for 22's distribution by only using the convolution on the joint distribution of XI and X2. Now, we use the convolution on the joint distribution of Z2 and X3 to obtain the distribution of Z3: But, we were interested in Y3, not Z3. SO,we perform a simple continuous inverse transformation of random variables using Z3 = 3Y3: 4. Compute its expectation: IE[Y3].
Solution to 4: Very easily we can apply the properties of expectations to determine again that IE[Y3] = 1. However, we can also determine this using the convolution formula. What we should have learned from part 2 was that all of the leading integration by parts terms will not matter as yGey = 0 for y = oo and for y = 0 for any k E N.

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