1 Solution to Problem 2.1

Transcription

1 Solution to Problem 2. I incorrectly worked this exercise instead of 2.2, so I decided to include the solution anyway. a) We have X Y /3, which is a - function. It maps the interval, ) where X lives) onto itself, so the support of the distribution) of Y is, ). f Y y) f ) X y /3 d Checking the integral, dy y/3 2y 5/3 y /3) 3 y 2/3 y y /3), < y <. y y /3) dy y y /3 ) dy [ 2 y2 3 ] 7 y7/3 [ 2 3 7] [ 7 6 ] [ ]. b) X Y 3)/ is -. The interval, ), which is the support of X, is mapped to 3, ) under Y X + 3, so the support of Y is 3, ). Computing the pdf for Y : d f Y y) f X y 3)/) y 3)/ dy Checking the integral, 3 y 7 exp 7y 3)/) 7 exp 7y 3)/), y > 3. 7 exp 7y 3)/) dy

2 , 7 exp 7x/) dx substitute x y 3) since the latter expression is the integral over its entire range of the exponential, mean /7, pdf. c) The function y gx) x 2 is - on x, ), and maps, ) onto, ), so the support of Y is, ). As X Y /2, Checking the integral, f Y y) f X y /2 ) d dy y/2 3y y /2) 2 2 y /2 5y /2 y /2) 2, < y <. 5y /2 y /2) 2 dy 5 5 y /2 2y /2 + y ) dy y /2 2y + y 3/2) dy [ y3/2 y ] 5 y5/2 [ ] + 2 [ ] y 2 Solution to Problem 2.2 We apply Theorem 2..5 in each case. 2

3 a) X Y /2 so dx /2)y /2 dy. Note that as x ranges over, ), x 2 also ranges over, ). f Y y) f ) X y /2 /2)y /2, < y <, /2)y /2, < y <. b) If Y log X, then X e Y and dx e y dy. As x ranges over, ), y ranges over, ), so we have f Y y) f X e y ) e y, y >, n + m + )! e ny e y) m e y, n!m! y >, n + m + )! e n+)y e y) m n!m! y >. c) If Y e X, then X log Y and dx y dy. Also, as x ranges over, ), y ranges over, ). Hence, f Y y) f X log y)y, y >, [ σ log y exp ] log 2 y)2 y, y >, 2σ2 [ log y exp ] log σ 2 y)2, y >. y 2σ2 3 Solution to Problem 2.9 The cdf for < x < 3 is Fx) x x y )/2 dy y/2 dy [ y 2 / ] x y x ) 2 /. 3

4 Thus, the cdf over the whole real line is if x <, Fx) x ) 2 / if x < 3, if 3 x. Note that Fx) is continuous at x as x ) 2 / at x, and also at x 3 as x ) 2 / at x 3. Thus, U X ) 2 / has a Unif,) distribution. Solution to Problem 2. a) Find E[X 2 ] directly must mean integrating against the N, ) pdf: x 2 e x2 /2 dx x xe x2 /2 ) dx. ) Applying integration by parts: udv uv vdu u x dv xe x2 /2 dx du dx v e x2 /2 we obtain x 2 e x2 /2 dx xe x2 /2, e x2 /2 dx + x e x2 /2 dx so we obtain E[X 2 ] after we put the / factor back in the pdf.

5 Now, using the pdf of Example 2..7: if Y X 2 then f Y y) 2 y [f X y) + f X y)] 2 y 2 e y/2 y /2 e y/2, y >. Now, E[Y ] y y /2 e y/2 dy y /2 e y/2 dy 2 y /2 e y/2 y + y /2 e y/2 dy, 2) where we have used integration by parts with u y /2 du 2 y /2 dy dv e y/2 dy v 2e y/2 Observe that the first term on the r.h.s. of 2) is, since y /2 at y and lim y y /2 e y/2. Also, the second term on the r.h.s. of 2) is since it is the integral of f Y y) over its whole range. b) The pdf of Y X is similar but easier) to the derivation in Example To give the details: assume y, F Y y) P[Y y] P[ X y] P[ y X y] P[ y < X y] since X is a cont. rv F X y) F X y). 5

6 Differentiating, we get f Y y) f X y) + f X y) 2f X y) since by symmetry f X y) f X y) recall X N, )). Thus, f Y y) 2 π e y2 /2, y >. We have E[Y ] 2 π 2 π 2 π. ye y2 /2 dy e w dw substitute w y 2 /2, dw ydy We have E[Y 2 ] E[X 2 ], so Var[Y ] E[Y 2 ] E[Y ]) 2 2/π Solution to Problem 2. a) There are various approaches to this problem under certain assumptions one can use integration by parts. However, the easiest approach by far is to use the following trick: Fx) P[X > x] x fy) dy. Plugging this into the expression on the r.h.s. of the proposed formula: Fx) dx x y 6 fy) dy dx fy) dxdy.

7 In the last expression, we simply switched the order of integration. Evaluating the double integral can be done by iterated integration either way: first over x or first over y. Note that we are integrating in the plane IR 2 ) over the region {x, y) : x y < }. Since fy) is a constant when integrating with respect to x, the last integral is yfy) dy E[X], since f ) is the pdf for X. b) A similar argument works here: so Fk) jk+ fj), [ Fk)] k fj) k jk+ j fj) j k jfj) j E[X]. 6 Solution to Problem 2.22 I guess I had too much spare time and prepared a solution for this exercise as well. a) Clear fx), as β >, x 2, and e x2 /2 >. Checking that it integrates to, β 3 π x2 x2 /β2 e dx 2 π y2 e y2 /2 dy 3) 2 y 2 e y2 /2 dy 7

8 y 2 e y2 /2 dy ). 5) where at 3), we substituted y 2x/β, at ) we used the symmetry of the N, ) pdf, and at 5) we used the fact that E[Y 2 ] when Y N, ), something we already proved back in Exercise 2.. b) Applying the same substitution as at 3), we have E[X] β 3 π x3 e x2 /β2 β π y3 e y2 /2 dy β y 2 e y2 /2 π 2β π y where at 6) we did an integration by parts with dx + 2β π ye y2 /2 dy 6) e w dw, 7) 2β π. 8) u y 2 dv ye y2 /2 dy du 2y dy v e y2 /2, and at 7) we substituted w y 2 /2. Proceeding on to the next moment, E[X 2 ] β 3 π x x2 /β2 e dx β 2 y e y2 /2 dy β2 y 3 e y2 /2 3β2 2 3β2 2. y y 2 e y2 /2 dy 8 + 3β2 y 2 e y2 /2 dy

9 Our integration by parts here was u y 3 dv ye y2 /2 dy du 3y 2 dy v e y2 /2 Thus, we have ) 2 Var[X] 3β2 2β 2 π [ 3 2 π] β β 2. 7 Solution to Problem 2.32 We have by the chain rule for differentiation, d d log Mt) dt Mt) dt Mt). Since M), we obtain that S ) M )/M) M ) E[X]. For the second derivative, we apply the rule for differentiating a quotient: so S 2) ) M)M ) M )) 2 M 2 ) S 2) t) d M t) dt Mt) Mt)M t) M t)) 2, M 2 t) M ) M )) 2 E[X 2 ] E[X]) 2 Var[X]. 8 Solution to Problem 2.33 a) Mgf for the Poissonλ): Mt) k e tkλk k! e λ e λ λe t ) k k! k e λ exp [ λe t] 9) exp [ λ e t )]. 9

10 At 9), we used the formula for the Taylor series for the exponential function. Since this converges everywhere, there are no restrictions on t, i.e. the mgf is finite everywhere. Using the trick from the previous problem, we have E[X] S ) d dt λ e t ) λe t t t λ, and Var[X] S ) d dt λet t λ. b) Mgf for the Geometricp): where we used the formula for summing a geometric series: Mt) e kt p) k p k p e t p) ) k k p e t p), ) k r k, r <. r The region for convergence r < translates into e t p) < t < log p). Note that log p) > for < p <, so the mgf is finite in a neighborhood of. Computing the moments: S t) d dt Substituting in t gives Also, [ log p log [ e t p) ]] e t p)/ [ e t p) ]. E[X] p)/ [ p)] p)/p. S t) d dt et p)/ [ e t p) ] d dt p)/ [ e t p) ] p)e t / [ e t p) ] 2,

11 and so evaluating this at t gives Var[X] p)/p 2. c) As noted in lecture, it is much easier to start with the N, ) case and then extend to Nµ, σ 2 ) using transformation formulae as in Theorem So, let Z N, ). Then M Z t) e t2 /2. Now, X σz + µ is Nµ, σ 2 ), and e tz e z2 /2 dz [ exp ] 2 z t)2 + t 2 /2 dz M X t) E[e tx ] E[exptσZ + tµ)] expµt)e[exptσz)] expµt)m Z σt) expµt) expσ 2 t 2 /2) expµt + σ 2 t 2 /2), as desired. Now the mean and variance computations are especially easy with the trick from Exercise 2.32: E[X] d µt + σ 2 t 2 /2 ) dt t and Var[X] d µ + σ 2 t ) dt µ + σ 2 t ) t µ, t σ 2.