APPM/MATH 4/5520 Solutions to Problem Set Two. = 2 y = y 2. e 1 2 x2 1 = 1. (g 1

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1 APPM/MATH 4/552 Solutions to Problem Set Two. Let Y X /X 2 and let Y 2 X 2. (We can select Y 2 to be anything but when dealing with a fraction for Y, it is usually convenient to set Y 2 to be the denominator.) We will find the joint distribution of Y and Y 2 and then integrate out Y 2 in order to get the marginal distribution for Y alone. we have y g (x, x 2 ) x /x 2 and y 2 g 2 (x, x 2 ) x 2 which gives us x g (y, y 2 ) y y 2 and x 2 g2 (y, y 2 ) y 2. The Jacobian of this transformation is x x y y 2 y J 2 y y 2 y y 2 Since X and X 2 are independent N(, ) random variables, we can get their joint pdf by multiplying two N(, ) pdfs: f X,X 2 (x, x 2 ) e 2 x2 e 2 x2 2 e 2 (x2 +x2 2 ). f Y,Y 2 (y, y 2 ) f X,X 2 (g (y, y 2 ), g 2 (y, y 2 )) J e 2 (y2 y2 2 +y2 2 ) y 2 (Note that there are no indicators anywhere because all variables go from to so there is no need to zero out the pdfs anywhere.) Now let s compute the marginal pdf for Y. f Y (y ) f Y,Y 2 (y, y 2 ) dy 2 y 2 e 2 y2 2 (+y2 ) dy 2 π y 2 e 2 y2 2 (+y2 ) dy 2 Method One: Do the integral. To do the integral, just think of + y 2 as a constant. In fact, let s call it c. f Y (y ) π y 2 e 2 cy2 2 dy2 With the substitution u y 2 2 (so du 2y 2 dy 2, and u still goes from to ), f Y (y ) π y 2 e 2 cy2 2 dy2 2y 2 e 2 cy2 2 dy2 e 2 cu du π c π(+y 2 )

2 and y goes from to. This is the pdf of the Cauchy distribution with α and β. Y X /X 2 has a Cauchy(α, β ) distribution. Method Two: Integrate without Integrating y 2 e 2 cy2 2 on (, ) looks like the pdf for a Weibull distribution. (The tip off is seeing e to a power to a power!) Specifically, it looks like the Weibull distribution with α, γ 2 and β 2/c which would be ( ) 2 2 y2 e (y 2/ 2/c) 2 cy 2 e 2 cy2 2. 2/c 2/c So f Y (y ) π y 2 e 2 cy2 2 dy2 π cy 2 e 2 cy2 2 dy2 }{{} πc π(+y 2 ) which again is the pdf of the Cauchy distribution with α and β. 2. We need the joint pdf of Y and Y 2. y g (x, x 2 ) x + x 2 and y 2 g 2 (x, x 2 ) x x 2 x g (y, y 2 ) 2 (y + y 2 ) and x 2 g 2 (y, y 2 ) 2 (y y 2 ) The Jacobian is J x y x y 2 y y 2 /2 /2 /2 /2 /2 Since X and X 2 are independent N(, ) random variables, we can get their joint pdf by multiplying two N(, ) pdfs: f X,X 2 (x, x 2 ) e 2 x2 e 2 x2 2 e 2 (x2 +x2 2 ). f Y,Y 2 (y, y 2 ) f X,X 2 (g (y, y 2 ), g 2 (y, y 2 )) J e 2[ 4 (y +y 2 ) (y y 2 ) 2 ] /2 4π e 8 (2y2 +2y2 2 ). (Note that there are no indicators anywhere because all variables go from to so there is no need to zero out the pdfs anywhere.)

3 Since this joint pdf factors into a y -part and y 2 -part (indicators, though not here, included), we have that Y and Y 2 independent. (The problem is done but, just for the record, both Y and Y 2 are N(, 2) random variables!) 3. First, note that the joint pdf of X and X 2 is f X,X 2 (x, x 2 ) indep f X (x ) f X2 (x 2 ) Γ(α) βα x α e βx I (, ) (x) βe βx 2 I (, ) (x 2 ) Γ(α) βα+ x α e β(x +x 2 ) I (, ) (x) I (, ) (x 2 ) Let Then y g (x, x 2 ) x 2 x + x 2 and y 2 g 2 (x, x 2 ) x + x 2. x g (y, y 2 ) y 2 y y 2 y 2 ( y ) and x 2 y y 2. The Jacobian of the transformation is x x y y 2 J y y 2 y 2 y y 2 y y y 2 y 2 ( y ) y 2. The new joint pdf is f Y,Y 2 (y, y 2 ) f X,X 2 (g (y, y 2 ), g 2 (y, y 2 )) J Γ(α) βα+ [y 2 ( y )] α e β[y 2( y )+y y 2 ] I (, ) (y 2 ( y )) I (, ) (y y 2 ) y 2 Γ(α) βα+ y α 2 ( y ) α e βy2 y 2 I (, ) (y 2 ( y )) I (, ) (y y 2 ). The indicators tells us that we need y 2 ( y ) > and y y 2 > to hold simultaneously. If you shade these regions in the y and y 2 plane, you will see that it corresponds to the infinite height rectangle with base sitting on the y -axis between and. Thus, the region is also described by the constraints < y < and < y 2 <. Since y 2 >, we can drop the absolute value on the y 2 in the joint pdf above and conclude that it is f Y,Y 2 (y, y 2 ) Γ(α) βα+ y α 2 ( y ) α e βy 2 I (,) (y ) I (, ) (y 2 ). Ultimately, we care about the pdf for Y only. In general, we must integrate out y 2 but here, they appear to be independent. Focusing on the y only, we have ( y ) α I (,) (y ). ()

4 While this is not a proper pdf, the constants from the joint pdf must sort themselves out correctly, leaving us with appropriate constants for the marginal pdfs. From (), we conclude that Y Beta(a, b α). 4. For maxes and mins, always start with cdfs! (Unless you want to plug and chug into a formula you have previously derived using cdfs!) (a) F X() (x) P (X () x) P (min(x, X 2,..., X n ) x) P (min(x, X 2,..., X n ) > x) P (X > x, X 2 > x,..., X n > x) indep P (X > x) P (X 2 > x) P (X n > x) ident [P (X > x)] n unif ( x) n f X() (x) d dx F X () (x) n( x) n ( ) n( x) n. To complete the pdf, we include the support for the minimum which is the same as the support for the original random variables: f X() (x) n( x) n I (,) (x). This is the pdf for the Beta distribution with parameters a and b n. X () Beta(, n) Some notes about matching up the constants for a pdf: (i) You don t have to do it! You found a pdf so it must integrate to. If that s the case and you try to change out that leading n for some other constant, it would no longer integrate to and wouldn t be a proper pdf. The Beta(, n) pdf is a proper pdf and looks like f(x) B(, n) ( x)n I (,) (x). Thus, we must have that n /B(, n)! (ii) Alternatively, you can use the facts that and that, for a positive integer n, So you could directly compute that B(a, b) Γ(a)Γ(b) Γ(a + b) Γ(n) (n )! B(, n) n. ( Excitement, not a factorial.)

5 (b) Now for the max... F X(n) (x) P (X (n) x) P (max(x, X 2,..., X n ) x) indep P (X x) P (X 2 x) P (X n x) ident [P (X x)] n unif x n f X(n) (x) d dx F X (n) (x) nx n. To complete the pdf, we include the support of X (n) : f X(n) (x) nx n I (,) (x). This is the pdf for the Beta distribution with parameters a n and b. X (n) Beta(n, ) 5. To find the mean (expected value) for X (), we need to know its distribution. First, note that the cdf for the P areto(γ) distribution is F (x) x γ ( + u) γ+ dx ( + x) γ. F X() (x) P (X () x) P (min(x, X 2,..., X n ) x) P (min(x, X 2,..., X n ) > x) iid [P (X > x)] n [ ] n (+x) γ (+x) γn This is the cdf of a Pareto distribution with parameter γn. You might recognize this because you used the P areto(γ) cdf in this problem. If you do not recognize it, then take the derivative to find the pdf and look in your table of distributions. we have X () P areto(γn). From the table of distributions, we see that the mean is as long as γn >. E[X () ] γn

6 6. This transformation from uniforms to normals is known as the Box-Muller Transformation and is usually used by software packages to simulate normal random variables. Consider X 2 + X2 2 : X 2 + X2 2 ( 2 ln U cos 2 (U 2 ) + ( 2 ln U 2 ) sin 2 (U 2 ) ( 2 ln U )(cos 2 (U 2 ) + sin 2 (U 2 )) ( 2 ln U ) 2 ln U To solve for U 2, I will use the fact that tan x sin x/ cos x, and consider the ratio X 2 /X : X 2 X 2 ln U sin(u 2 ) 2 ln U cos(u 2 ) we have that sin(u 2) cos(2πu 2 ) tan(u 2) U g (X, X 2 ) e 2 (X2 +X2 2 ) and U 2 g 2 (X, X 2 ) tan and the Jacobian of the transformation is u u x J x u 2 u 2 e 2 (x2 +x2 2 ) x 2 e 2 (x2 +x2 2 ) ( ) ( ) x x x 2 x 2 +x2 2 x 2 x 2 +(x 2 /x ) 2 x e 2 (x2 +x2 2 ) x 2 e 2 (x2 +x2 2 ) x x 2 +x2 2 Therefore, the joint density of X and X 2 is given by f X,X 2 (x, x 2 ) f U,U 2 (g (x, x 2 ), g 2 (x, x 2 )) J +(x 2 /x ) 2 e 2 (x2 +x2 2 ). ( ) X2 I (,) (e 2 (x2 +x2)) ( ( )) 2 I (,) tan x 2 x e 2 (x2 +x2 2 ) Here begins a long note about indicators... brace yourself. Note that the first indicator is always since e 2 (x2 +x2 2 ) lives between and. (Don t be alarmed about the idea that it could equal. That will happen when both x and x 2 are zero. Since they are continuous random variables, this will happen with probability zero. For similar reasons, you could have started with your uniforms on [, ] as opposed to (, ) since those single endpoints don t matter for a continuous random variable.) Since the first indicator is always, we can drop it. Now, we are left with only one indicator that takes the value whenever < tan (x 2 /x ) <. By examination of arctan, this appears to only be true for x 2 /x >, which leads us to quadrants and 3 of the (x, x 2 ) plane. This does not seem to have us heading in the right direction for independent standard normal random variables! The problem here is with the definition of the arctan. y tan(x) is not invertible until we restrict its domain. It is usually restricted to ( π/2, π/2) but doesn t have to be. We may X

7 invert it on different regions. In this problem, the restriction ends up restricting possible values for x and x 2! Indeed, at a previous point in this solution, we had that x 2 x tan(u 2 ). Here, the right-hand side is taking on values from to so the left-hand side should be able to take on these values as well. It appears at this point that x and x 2 can both take on values from to. It was not until we applied the arctan that we got some restriction. Indeed, if one looks at the original definitions of x and x 2, x 2 ln u cos(u 2 ) x 2 2 ln u sin(u 2 ) it is easy to see that any (x, x 2 ) pair is possible. Since this region (the entire plane!) is rectangular, we have that the joint pdf for X and X 2 is f X,X 2 (x, x 2 ) e 2 (x2 +x 2) I (, ) (x ) I (, ) (x 2 ) e 2 x2 I(, ) (x ) e 2 x2 2 I(, ) (x 2 ). Since the joint density factors, we see that X and X 2 are independent. Furthermore, from the form of the desity, we see that they are N(, ) random variables! 7. The pdf for X is f(x) ( p) x p I {,,2,...}. The moment generating function is therefore M(t) E[e tx ] x etx P (X x) x e tx ( p) x p p x [e t ( p)] x This looks like a nice geometric sum of the form n r n r if r <. (Note that the sum blows up if we do not have r <.) For our particular problem, the sum is geometric with r e t ( p). converge for r e t ( p) e t ( p) <. In terms of a range of acceptable t values, this means that we need ( ) t < ln ln( p). p The sum will only we will finish off our moment generating function under this assumtion for t. It is M(t) p [e t ( p)] x p e x t ( p)

8 which, again, is valid for t < ln( p). This matches the mgf given in the table of distributions! Yay! 8. There are slicker ways to do this problem. We will show a method that is in line with the basic concepts we have been talking about so far in class. It is usually convenient to think of distributions of mins and or maxes in terms of cdfs first and then to take derivatives to get pdfs. This problem is no exception. We will consider the joint cdf: P (X () x, X (n) y) y x f X(),X (n) (u, v) du dv. Note that, at this point, we do not have to be careful about the relative position of the arguments x and y for this to be true. Indicators in f X(),X (n) (u, v) will take care of that in the end! However, we can get the joint pdf back from the joint cdf by taking derivatives with respect to x and y. Note that P (X () x, X (n) y) if y x. assume that x < y. Following the ideas from the marginal distributions of X () and X (n), you might want to rewrite the event {X () x, X (n) y} in terms of X, X 2,..., X n. This is difficult to do straight away. However, because we have that {X (n) y} {X () x, X (n) y} {X () > x, X (n) y} with the events on the right-hand side being disjoint. Thus, we have that and therefore that Term : P (X (n) y) P (X () x, X (n) y) + P (X () > x, X (n) y) P (X () x, X (n) y) P (X (n) y) P (X () > x, X (n) y) }{{}}{{} 2 P (X (n) y) P (max(x, X 2,..., X n ) y) P (X y, X 2 x,..., X n x) indep P (X x) P (X 2 x) P (X n x) ident [P (X x)] n [F (x)] n where F is the cdf of any one of the original X, X 2,..., X n. Term 2 : Note that the event {X () > x, X (n) y} is equivalent to the event that all of the indinividual X i are between x and y. (Think about it!)

9 Thus, we have P (X () > x, X (n) y) P (x < X y, x < X 2 y,..., x < X n y). By independence of X, X 2,..., X n, this is equal to P (x < X y) P (x < X 2 y) P (x < X n y). By identicalness this is [x < X y] n [F (x) F (y)] n where F is the cdf of any one of the original X, X 2,..., X n. Putting it all together, we have P (X () x, X (n) y) [F (x)] n [F (x) F (y)] n. Taking the derivative with respect to y first yields (you could do x first instead) n[f (x) F (y)] n f(x) n[f (x) F (y)] n f(x). With respect to x now yields n(n )f(x)[f (x) F (y)] n 2 ( f(y)) n(n )f(x)f(y)[f (x) F (y)] n 2. Recall that this only hold for x < y. The joint pdf is otherwise. the final answer is { n(n )f(x)f(y)[f (x) F (y)] n 2, x < y f X(),X (n) (x, y), otherwise.