4. Suppose fX;Y (x; y) = ke^-(x+y); 0  < equal to x < infinity ; 0 < equal to

Question

4. Suppose fX;Y (x; y) = ke^-(x+y); 0  < equal to x < infinity ; 0 < equal to y < infinity: (a) What is the value of k? (b) Are X and Y independent? Why or why not? (c) Determine FY (3). (d) Determine FX;Y (5,3). (e) Find the marginal pdf of X. What is the name of this distribution?

Explanation / Answer

Solutions and Comments for Assignment 2 1. Suppose that f : Rn ! Rn is continuously di®erentiable and one-to-one, so that its inverse g is de¯ned on the range of f. Suppose also that f0(0) is not invertible. Prove that g cannot be di®erentiable at f(0). The idea here is that f should be a function like f(x) = x3 with f0(0) = 0. Sure enough, its inverse g(y) = y1=3 is not di®erentiable at y = 0. This problem asks you to prove that this is true in general. To prove that by contradiction suppose that the inverse function is di®erentiable at y = f(0). Our de¯nition of di®erentiable assumes that g is de¯ned on B(f(0); r) for some r > 0. So I will assume this without trying to prove that the range of f contains B(f(0); r): if it doesn't, we have already proven that g is not di®erentiable at f(0). Then we have g(f(x)) = x for x 2 B(0; s) for some s > 0, and the chain rule implies g0(f(0))f0(0) = I which says that f0(0), which was assumed to be not invertible, has an inverse. This contradiction is all you need. 2. Let f : Rn ! R be di®erentiable. If for all t > 0 one has f(tx) = tf (x) for all x 2 Rn, show that f(x) = rf(0) ¢ x. However, if you take any function g which is di®erentiable at all points on the sphere fjxj = 1g, and de¯ne f(0) = 0 and f(x) = jxjg(x=jxj) for x 6= 0, f will have (one-sided) directional derivatives in all directions at x = 0. This is quite easy. Just let x be any ¯xed point in Rn, di®erentiate both sides of f(tx) = tf (x) with respect to t, and take the limit as t ! 0. The left hand side requires using the chain rule. The example was supposed to illustrate the way that this result fails if you do not assume di®erentiability of f: in the example f is di®erentiable at all points except x = 0, and even at x = 0 it has directional derivatives in all directions, Dvf(0) = g(v). 3. (This one is from Edwards, p.89) Let w = f(x; y; z) and z = g(x; y). Then by the chain rule @w @x = @w @x @x @x + @w @y @y @x + @w @z @z @x : (¤) Simplifying this, since @x @x = 1 and @y @x = 0, you can conclude @w @z @z @x = 0. However, for w = x+y +z and z = g(x; y) = x+y we have @w @z = @z @x = 1 and (¤) gives 0=1. Where exactly is the mistake in this? This is just a simple mistake: the functions w(x; y; z) and w(x; y; g(x; y)) are both denoted by w. This would be OK if one did not go on to identify the x-derivative of the ¯rst: @w @x (x; y; z) with the x-derivative of the second: @w @x (x; y; g(x; y)) + @w @z (x; y; g(x; y)) @g @x (x; y): 2 4. (Basic S'07) Suppose that the real-valued functions ffng are twice continuously di®erentiable on [0; 1] and satisfy lim n!1 fn(x) = f(x); jf0n (x)j · 1; and jf00 n (x)j · 1 for all x 2 [0; 1] and n ¸ 1. Prove that f is continuously di®erentiable on [0; 1]. The derivatives ff0n g are an equicontinuous family, and the Arzela-Ascoli theorem gives you a subsequence ff0n k g1k =1 converging uniformly to g 2 C([0; 1]). For the rest use fnk (x) ¡ fnk (0) = Z x 0 f0n k (t)dt: When you take the limit as k ! 1, the uniform convergence of the f0n k implies the convergence of the integrals. Hence u(x) ¡ u(0) = Z x 0 g(t)dt; and u is continuously di®erentiable. 5. Consider the following function on R2: f(x; y) = ½ x2 + y2; if x and y are rational numbers 0 elsewhere At what points is f continuous? At what points is f di®erentiable? Since the points with rational coordinates are dense and the points with irrational coordinates are also dense, f is only continuous at (x,y)=(0,0). However, directly from the de¯nition, you can see that it is di®erentiable there too with rf(0; 0) = (0; 0). 6. (Basic S '06) Let W be the set of continuous real-valued functions on [0,1] satisfying jf(x) ¡ f(y)j · jx ¡ yj and Z 1 0 (f(x))2dx · 1: a) Show that there is an M such that jf(x)j · M for all x 2 [0; 1] and all f 2 W. b) Show that W is a compact subset of the set of continuous real-valued functions on [0,1], considered as a metric space with d(f; g) = maxx2[0;1] jf(x) ¡ g(x)j: This problem is the same as problem 5 on the ¯rst assignment! In a desper- ate attempt to justify this glitch I point out that this prepares for the frequently rRecycled problems on the Basic Exam. The recycling here consisted of replacing 1 0 f(x)dx = 1 by R 1 0 (f(x))2dx · 1. That really makes no di®erence, since, com- bined with jf(x)¡f(y)j · jx¡yj, it still implies that jf(x)j · 2. That was the use of R 1 0 f(x)dx = 1 in the solution of problem 5. 7. Suppose that F is continuously di®erentiable on R3, F(x0; y0; z0) = 0 and each component of rF(x0; y0; z0) is nonzero. The the implicit function theorem gives 3 three functions f, g and h such that z = f(x; y), y = h(x; z) and x = g(y; z) all de¯ne the surface F(x; y; z) = 0 near (x0; y0; z0). Prove the relation @g @y (y; z)@h @z (x; z)@f @x (x; y) = ¡1: This is a computation with the chain rule. Di®erentiate F(x; y; f(x; y)) ´ 0 with respect to x, F(x; h(x; z); z) ´ 0 with respect to z, and F(g(y; z); y; z) ´ 0 with respect to y: @F @x (x; y; f(x; y)) + @F @z (x; y; f(x; y))@f @x (x; y) = 0 @F @z (x; h(x; z); z) + @F @y (x; h(x; z); z)@h @z (x; z) = 0 @F @y (g(y; z); y; z) + @F @x (g(y; z); y; z)@g @y (y; z) = 0 When you sort out the results, you get the identity. One important point: by the uniqueness in the implicit function theorem (x; y; f(x; y)), (x; h(x; y)x; z) and (g(y; z); y; z) are all the same point when (x; y; z) is in a su±ciently small ball centered at (x0; y0; z0). 8. (Basic S'02) Suppose f : R3 ! R is continuously di®erentiable with rf(0; 0; 0) 6= 0. Show that there exist continuously di®erentiable g and h de¯ned near (0,0,0) such that F(x; y; z) = (f(x; y; z); g(x; y; z); h(x; y; z)) is one-to-one on some neigh- borhood of (0,0,0). This problem may seem odd, but it is very easy: if @f @x (0; 0; 0) 6= 0, you can choose vectors v and w so that frf(0; 0; 0); v;w; g is a basis for R3. Then take g = v¢(x; y; z) and h = w¢(x; y; z). De¯ning F(x; y; z) = (f(x; y; z); g(x; y; z); h(x; y; z)), the Jacobian matrix F0(0; 0; 0) will be the matrix with rows rf(0; 0; 0), v and w. Thus the inverse function theorem applies and shows that F is one-to-one on a ball centered at (0; 0; 0).

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