MATH 217A HOMEWORK. P (A i A j ). First, the basis case. We make a union disjoint as follows: P (A B) = P (A) + P (A c B)

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1 MATH 217A HOMEWOK EIN PEASE 1. (Chap. 1, Problem 2. (a Let (, Σ, P be a probability space and {A i, 1 i n} Σ, n 2. Prove that P A i n P (A i P (A i A j + P (A i A j A k... + ( 1 n 1 P A i n P (A i P (A i A j. First, the basis case. We make a union disjoint as follows: A B A (A c B. Thus P (A B P (A + P (A c B. (1 Similarly, we can write B (A B (A c B P (B P (A B + P (A c B P (B P (A B P (A c B, which we plug into (1 to get P (A B P (A + P (A c B P (A + P (B P (A B. Now we proceed by induction. Assume n P A i P (A i P (A i A j + P (A i A j A k... + ( 1 n 1 P A i. 1

2 2 EIN PEASE P Using the basis case and then the inductive hypothesis gives +1 A i P A i A n+1 ( n P A i + P (A n+1 P A i A n+1 n+1 P (A i P (A i A j + P (A i A j A k ( n... + ( 1 n 1 P A i P A i A n+1. (2 Using the inductive hypothesis again, the last term in (2 becomes ( n P A i A n+1 P (A i A n+1 n+1 P (A i A n+1 + by distribution P (A i A j A n+1 P (A i A j A k A n ( 1 n 1 P A i A n+1 We plug this back into (2, and get, for example, n+1 P (A i A j P (A i A n+1 +1 P (A i A j. Similarly, collecting like terms in the other sums (i.e., terms with the same number of A j s getting unioned together and rearranging gives P +1 A i n P (A i P (A i A j + P (A i A j A k... + ( 1 n 1 P A i, as desired.

3 MATH 217A HOMEWOK 3 2. (Chap. 1, Problem 3. (a Let {X n, n 1} be a sequence of random variables on a probability space (, Σ, P. Show that ( P Xn X X n X E ( Let X n we can say ( Xn X E 1 + P X, i.e., P [ ε] A n,ε : { < ε}, 1 + dp A n,ε 1 + dp + ε < 1 + ε dp + 1dP A n,ε A n,ε \N where P (N 0. But then lim E A C n,ε 0. Now if we define A C n,ε ε 1 + ε dp + P [ ε] ε 1 + ε dp + 0 ( Xn X 1 + for any ε > 0, which shows lim E ( Now assume lim E Now P [ ε] ε dp ε dp ε ε 1 + ε dp ( Xn X 1+ X ( n X Xn X 1+ X n X A n : { ε}. χ An dp lim Define 1 + χ A n dp 1 + dp

4 4 EIN PEASE (b Verify that lim 1 + χ A n dp lim 1 + dp ( Xn X lim E d(x, Y : E ( X Y 1 + X Y defines a metric on the space of random variables L 0, and that L 0 is an algebra. (i Positivity. Clearly, d(x, X E Then define A {X Y } and A n { X Y 1 n }. Now we have X Y d(x, Y A 1 + X Y dp X Y A n 1 + X Y dp 1/n A n 1 + 1/n dp n n + 1 P (A n > 0 for P (A n > 0. ( X X E(0 0. So let X Y. 1+ X X So X Y implies there is some n for which P (A n > 0, in which case d(x, Y > 0. ( ( (ii Symmetry. d(x, Y E X Y E Y X d(y, X. 1+ X Y 1+ Y X (iii Triangle inequality. Consider the function f : + [0, 1] by f(x Taking derivatives of this function shows that it is concave increasing with slope less than 1 for all x > 0. Alternatively, see Lemma 1 in Problem 2. This gives f(a + b f(a + f(b immediately, for a, b 0. Using a X Y and b Y Z, f(a + b X Y + Y Z 1 + X Y + Y Z x. 1+x

5 Since MATH 217A HOMEWOK 5 X Y Y Z X Y 1 + Y Z f(a + f(b. X Z X Y + Y Z 1 + X Z 1 + X Y + Y Z by the triangle inequality, X Z d(x, Z 1 + X Z dp X Y 1 + X Y dp + d(x, Y + d(y, Z. Y Z 1 + Y Z dp To see that L 0 is an algebra, we make some basic observations, namely: (i A sum of measurable functions is again measurable. (ii The pointwise product of measurable functions is again measurable. (iii Any scalar multiple of a measurable function is again measurable. Pointwise multiplication is associative, even commutative. identity f(x 0 and unit g(x 1. Also, we have the 3. (Chap. 2, Problem 2. (a Let φ : + be a continuous function such that φ is increasing and convex on +, and with φ(0 0 and φ( x x. Also, assume φ satisfies φ(2x cφ(x for x 0, for some 0 < c <. Let X i :, i 1, 2 be two random variables on (, Σ, P. If E(φ(X i <, i 1, 2, then verify E(φ(X 1 + X 2 <. Show the converse is also true if the X i are independent. ( Since φ increasing implies φ is order-preserving, we bound E(φ(X 1 + X 2 as follows: E(φ(X 1 + X 2 φ(x 1 + X 2 dp φ(2x 1 dp + φ(2x 2 dp φ increasing {X 1 X 2 } {X 2 >X 1 } E(φ(2X 1 + E(φ(2X 1 ce(φ(2x 1 + ce(φ(x 1 P is monotone given <

6 6 EIN PEASE ( Now we take X 1, X 2 to be independent. Then with A n { X 2 n}, E(φ(X 1 + X 2 E(φ( X 1 + X 2 E (φ ( X 1 X 2 φ( x φ(x a + b a b E (φ ( X 1 X 2 φ( x φ(x φ ( X 1 X 2 dp def of E φ ( X 1 X 2 dp A n + φ ( X 1 X 2 dp A n A c n A c n φ ( X 1 n dp + 0 def of A n A n E (φ ( X 1 n χ An def of E E (φ ( X 1 n P (A n. independence Now we take note of two things. First, A n P (A n 1, so we may assume 0 < P (A n 1 and concern ourselves just with the other factor. Second, E (φ ( X 1 n φ ( x 1 n df X (x φ ( x 1 df X (x E (φ ( X 1 by FLoP. (Translation doesn t matter when we integrate over all of. Thus E(φ(X 1 + X 2 E (φ ( X 1 P (A n E(φ( X 1 E(φ(X 1 Since a similar procedure may be used to bound E(φ(X 2, we have E(φ(X 1, E(φ(X 2 <.

7 MATH 217A HOMEWOK 7 (b Let φ : + be a continuous function such that φ is increasing and concave on +, and with φ(0 0 and φ( x x. Let X i :, i 1, 2 be two random variables on (, Σ, P. If E(φ(X i <, i 1, 2, then verify E(φ(X 1 + X 2 <. Show the converse is also true if the X i are independent. ( First we prove the following lemma. Lemma 1. If φ is concave on +, then for any x, y > 0 we have φ(x + y φ(x + φ(y. Proof. Method 1. Since φ is concave, it is absolutely continuous on any open interval and hence may be represented as the integral of its derivative. Thus we may write φ(x + y x+y 0 x 0 φ(x + φ(x + φ (t dt φ (t dt + y 0 y 0 x+y x φ (t + x dt φ (t dt φ (t dt CoV φ(x + φ(y, FToC φ decreasing where the inequality is due to the fact that φ is decreasing whenever φ is concave. Proof. Method 2. Wlog, take 0 < x < y. Then x < y < x + y, so y αx + (1 α(x + y for α x y Then concavity means φ(y φ(αx + (1 α(x + y αφ(x + (1 αφ(x + y (0, 1. φ(x + φ(y φ(x + αφ(x + φ(x + y αφ(x + y. So it remains to show But this is just equivalent to φ(x + αφ(x αφ(x + y 0. φ(x + αφ(x αφ(x + y φ(x + xφ(x x φ(x + y y y (x + yφ(x xφ(x + y φ(x x φ(x + y (x + y,

8 8 EIN PEASE which is another form of the definition of concavity; the decreasing secants: s < t < u f(t f(s t s f(u f(s u s f(u f(t, u t with s 0, t x, u x + y. Hence, E (φ (X 1 + X 2 E (φ ( X 1 + X 2 φ( x φ(x E (φ ( X 1 + X 2 -ineq E (φ ( X 1 + φ ( X 2 by Lemma E (φ ( X 1 + E (φ ( X 2 linearity E (φ(x 1 + E (φ(x 2 φ( x φ(x < ( The converse here goes through exactly as it did in the previous case. 4. (Chap. 2, Problem 3. Let X 1, X 2 : be independent with E(X 1 0. Again, take φ : + to be a continuous function which is increasing and convex on +, and satisfies φ(0 0 and φ( x x. Prove that E(φ(X 1 +X 2 < implies E(φ(X 2 E(φ(X 1 +X 2. If E(X 2 0 is also assumed, prove E(φ(X i E(φ(X 1 + X 2. We write φ(x 2 φ(0 + x 2 φ (E(X 1 + x 2 E(X 1 0 φ (E(X 1 + x 2 linearity E (φ(x 1 + x 2 φ(x 1 + x 2 df X1. Now we integrate both side with respect to df X2, as follows: φ(x 2 df X2 E(φ(X 2 φ(x 1 + x 2 df X1 df X2 φ(x 1 + x 2 df X1 +X 2 E(φ(X 2 E(φ(X 1 + X 2. Jensen s ineq FLoP For the case E(X 2 0, the identical technique may be applied. independence

9 MATH 217A HOMEWOK 9 If X 1, X 2 0, the problem becomes easy. first we note that the Lemma proven in the previous problem will also work for convex functions, with the inequality reversed: φ(x + y φ(x + φ(y. Then E(φ(X 1 + X 2 E(φ(X 1 + φ(x 2 The other case follows similarly. Lemma E(φ(X 1 + E(φ(X 2 linearity φ(e(x 1 + E(φ(X 2 Jensen s φ(0 + E(φ(X 2 E(X 1 0 E(φ(X 2. φ(0 0

d(x n, x) d(x n, x nk ) + d(x nk, x) where we chose any fixed k > N

d(x n, x) d(x n, x nk ) + d(x nk, x) where we chose any fixed k > N Problem 1. Let f : A R R have the property that for every x A, there exists ɛ > 0 such that f(t) > ɛ if t (x ɛ, x + ɛ) A. If the set A is compact, prove there exists c > 0 such that f(x) > c for all x