Recursive Estimation

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1 Recursive Estimation Raffaello D Andrea Spring 28 Problem Set : Probability Review Last updated: March 6, 28 Notes: Notation: Unless otherwise noted, x, y, and z denote random variables, p x denotes the probability density function of x, and p x y denotes the conditional probability density function of x conditioned on y. Note that shorthand (as introduced in Lecture and 2) and longhand notation is used. The expected value of x and its variance is denoted by E [x] and Var [x], and Pr (Z) denotes the probability that the event Z occurs. A normally distributed random variable x with mean µ and variance σ 2 is denoted by x N ( µ, σ 2). Please report any errors found in this problem set to the teaching assistants (hofermat@ethz.ch or csferrazza@ethz.ch).

2 Problem Set Problem Let x be a discrete random variable (DRV) with sample space X = {,,, 2} and probability density function (PDF) p(x) = ax 2. a) Determine a such that p(x) is a valid PDF. b) Calculate the probability Pr (x = ). c) Calculate the probability Pr ( x ). d) Calculate the probability Pr ( x > ). e) Calculate the expected value E [x]. f) Calculate the variance Var [x]. Problem 2 Let x be a continuous random variable (CRV) with sample space X = [, 2] and probability density function (PDF) p(x) = ax 2. a) Determine a such that p(x) is a valid PDF. b) Calculate the probability Pr (x = ). c) Calculate the probability Pr ( x ). d) Calculate the probability Pr ( x > ). e) Calculate the expected value E [x]. f) Calculate the variance Var [x]. Problem 3 Let x and y be continuous random variables (CRVs) with sample space X = [, 2], Y = [, ] and joint PDF p(x, y) = 2 (x + y)2. a) Determine p x. b) Determine p y. c) Determine p x y. d) Determine p y x. e) Compute Pr ( x > y = 2), the probability that x is greater than given that y is.5. Problem 4 Prove p(x y) = p(x) p(y x) = p(y). 2

3 Problem 5 Prove p(x y, z) = p(x z) p(x, y z) = p(x z) p(y z). Problem 6 Come up with an example where p(x, y z) = p(x z) p(y z) but p(x, y) p(x) p(y) for continuous random variables x, y, and z. Problem 7 Come up with an example where p(x, y z) = p(x z) p(y z) but p(x, y) p(x) p(y) for discrete random variables x, y, and z. Problem 8 Let y be a scalar continuous random variable with probability density function p y, and g be a function transforming y to a new variable x by x = g(y). Derive the change of variables formula, i.e. the equation for the probability density function p x, under the assumption that g is continuously differentiable with dg(y) dy < y. Problem 9 Let x X be a scalar valued continuous random variable with probability density function p x. Prove the law of the unconscious statistician, that is, for a function g show that the expected value of y = g(x) is given by E [y] = y X g( x)p x ( x) d x. You may consider the special case where g is a continuously differentiable, strictly monotonically increasing function. Problem Prove E x [ax + b] = ae [x] + b, where a and b are constants. Problem Let x and y be scalar random variables. Prove that, if x and y are independent, then for any functions g and h, E x,y [g(x)h(y)] = E x [g(x)] E y [h(y)]. Problem 2 From above, it follows that if x and y are independent, then E [xy] = E [x] E [y]. Is the converse true, that is, if E [xy] = E [x] E [y], does it imply that x and y are independent? If yes, prove it. If no, find a counter-example. 3

4 Problem 3 Let x X and y Y be independent, scalar valued discrete random variables. Let z = x + y. Prove that p z ( z) = ȳ Y p x ( z ȳ) p y (ȳ) = x X p x ( x) p y ( z x), that is, the probability density function is the convolution of the individual probability density functions. Problem 4 (adapted from S.M. Ross, Introduction to Probability Models, 2) Let x, x 2,..., x n be independent random variables, each having a uniform distribution over [, ]. Let the random variable m be defined as the maximum of x,..., x n, that is, m = max{x, x 2,..., x n }. Show that the cumulative distribution function of m, F m ( ), is given by F m ( m) = m n, m. What is the probability density function of m? Problem 5 (adapted from S.M. Ross, Introduction to Probability Models, 2) Prove that E [ x 2] (E [x]) 2. When does the statement hold with equality? Problem 6 (adapted from S.M. Ross, Introduction to Probability Models, 985) Suppose that x and y are independent scalar continuous random variables. Show that Pr (x y) = F x (ȳ)p y (ȳ) dȳ. Problem 7 (adapted from S.M. Ross, Introduction to Probability Models, 2) Let x and y be independent random variables with means µ x and µ y and variances σ 2 x and σ 2 y, respectively. Show that Var x,y [xy] = σ2 xσ 2 y + µ 2 yσ 2 x + µ 2 xσ 2 y. Problem 8 a) Use the method presented in class to obtain a sample x of the exponential distribution { λe λ x x ˆp x ( x) = x < (λ > ) from a given sample u of a uniform distribution on [, ]. b) Let x and y be CRVs with (x, y) {( x, ȳ) R 2 x <, x ȳ < } and joint PDF { λ λ 2 e λ x e λ 2(ȳ x) x <, x ȳ < ˆp xy ( x, ȳ) = otherwise, with < λ 2 < λ. From two given samples u and u 2 of a uniform distribution on [, ], compute a sample (x s, y s ) of x and y using the method presented in class. 4

5 Additional Problems The following problems are not directly related to the topics covered in the lectures. We will not assume that you know these problem s specific results by heart, for example Chebychev s inequality. However, these problems let you further practice the application of the fundamental probability concepts discussed in the first two lectures. Problem 9 Let x and y be binary random variables: x, y {, }. Prove that p xy ( x, ȳ) p x ( x) + p y (ȳ), which is called Bonferroni s inequality. Problem 2 Let x be a scalar continuous random variable that takes on only non-negative values. Prove that, for a >, Pr (x a) E [x] a, which is called Markov s inequality. Problem 2 Let x be a scalar continuous random variable. Let x = E [x], σ 2 = Var [x]. Prove that for any k >, Pr ( x x k) σ2 k 2, which is called Chebyshev s inequality. Problem 22 (adapted from S.M. Ross, Introduction to Probability Models, 2) Use Chebyshev s inequality to prove the weak law of large numbers. Namely, if x, x 2,... are independent and identically distributed with mean µ and variance σ 2 then, for any ε >, ( ) x + x x n Pr µ n > ε as n. Problem 23 (adapted from S.M. Ross, Introduction to Probability Models, 2) Suppose that x is a random variable with mean and variance 5. What can we say about Pr (5 < x < 5)? Problem 24 Exponential random variables, such as the one in Problem 8, are used to model failures of components that do not wear, for example solid state electronics. This is based on their property Pr (x > s + t x > t) = Pr (x > s) s, t. A random variable with this property is said to be memoryless. Prove that a random variable with an exponential distribution satisfies this property. 5

6 Sample solutions Problem a) We require x X p(x) =. We find b) Pr (x = ) = p(x = ) = 6. c) Pr ( x ) = x= p(x) = 3. 2 x= ax 2 = 6a and therefore, a = 6. d) From above: Pr ( x > ) = 3 = 2 3, which is equal to Pr (x = 2). e) E [x] = xp(x) = = 4 3. x X f) Var [x] = E [ (x E [x]) 2] = E = E [ x 2] E [x] 2 = x X [ x 2 2xE [x] + E [x] 2] = E [ x 2] 2E [xe [x]] + E [E [x] 2] x 2 p(x) 6 9 = ( ) = = 9. Problem 2 a) We require x X p(x) dx =. We find 2 ax 2 dx = a [ ] x = 3a and therefore, a = 3. b) Pr (x = ) = c) Pr ( x ) = p(x) dx =, since the integration interval has zero length. p(x) dx = 3 [ ] x 3 3 = 2 9. d) From above: Pr ( x > ) = 2 9 = 7 9. e) E [x] = X xp(x) dx = 3 2 x 3 dx = 3 [ ] x = 5 4. f) From Problem f: Var [x] = E [ x 2] E [x] 2 = = 2 3 x 4 dx 25 6 = 3 2 [ ] x = = 5 8. x 2 p(x) dx 25 6 Problem 3 a) We apply marginalization: p(x) = y Y b) We apply marginalization: p(y) = x X c) We apply conditioning: p(x y) = p(x,y) p(y) = (x+y)2 3(+y+y 2 ). d) We apply conditioning: p(y x) = p(x,y) p(x) = 3(x+y)2 (+3x+3x 2 ). p(x, y) dy = 2 ( 3 + x + x2). p(x, y) dx = 6 ( + y + y 2 ). e) 2 p(x y =.5) = ( ) x =

7 Problem 4 From the definition of conditional probability, we have p(xy) = p(x y) p(y) and p(xy) = p(y x) p(x) and therefore p(x y) p(y) = p(y x) p(x). () We can assume that p(x) and p(y) ; otherwise p(y x) and p(x y), respectively, are not defined, and the statement under consideration does not make sense. We will check both directions of the equivalence statement: p(x y) = p(x) p(y x) = p(y) () = p(x) p(y) = p(y x) p(x) p(x) = p(y) = p(y x) () = p(x y) p(y) = p(y) p(x) p(y) = p(x) = p(x y) Therefore, p(x y) = p(x) p(y x) = p(y). Problem 5 Using p(x, y z) = p(x y, z) p(y z) (2) both directions of the equivalence statement can be shown as follows: p(x y, z) = p(x z) p(x, y z) = p(x z) p(y z) (2) = p(x, y z) = p(x z) p(y z) (2) = p(x y, z) p(y z) = p(x z) p(y z) = p(x y, z) = p(x z) since p(y z), otherwise p(x y, z) is not defined. Problem 6 Let x, y, z [, ] be continuous random variables. Consider the joint PDF p(x, y, z) = c g x (x, z) g y (y, z) where g x and g y necessarily PDFs) and c is a constant such that are functions (not p(x, y, z) dx dy dz = Then with p(x, y z) = p(x, y, z) p(z) = c g x(x, z) g y (y, z), c G x (z) G y (z) G x (z) = g x (x, z) dx, G y (z) = g y (y, z) dy 7

8 since p(z) = c g x (x, z) g y (y, z) dx dy = c G x (z) g y (y, z) dy = c G x (z) G y (z). Furthermore, p(x z) = p(x, z) p(z) = p(x, y, z) dy p(z) = c g x(x, z)g y (z) c G x (z)g y (z) = g x(x, z) G x (z) and, by analogy, p(y z) = g y(y, z) G y (z). Continuing the above, p(x, y z) = g x(x, z)g y (y, z) G x (z)g y (z) = p(x z) p(y z), that is, x and y are conditionally independent. Now, let g x (x, z) = x + z and g y (y, z) = y + z. Then, p(x, y) = = c p(x, y, z) dz = c [ 3 z3 + 2 (x + y)z2 + xyz (x + z)(y + z) dz = c ] z= z= z 2 + (x + y)z + xy dz ( = c 3 + ) (x + y) + xy 2 p(x) = p(y) = Therefore, p(x, y) dy = c p(x, y) dx = c x + ( y + xy dy = c x ) ( 2 x = c x + 7 ) 2 ( y + 7 ) 2 (by symmetry). ( 49 p(x) p(y) = c ) (x + y) + xy p(x, y), 2 that is, x and y are not independent. Problem 7 As in Problem 6, let p(x, y, z) = c g x (x, z)g y (y, z). This ensures that p(x, y z) = p(x z) p(y z). Let x, y, z {, }. 8

9 Define g x and g y by value tables: x z g x (x, z) Compute p(x, y), p(x), p(y): y z g y (y, z) x y z p(x, y, z) = 2 g x(x, z)g y (y, z).5.5 x y p(x, y).5.5 From this we see p(x, y) p(x) p(y). x p(x).5.5 y p(y).5.5 x y p(x) p(y) Remark: For verification, one may also compute p(x, y z), p(x z), p(y z) in order to verify p(x, y z) = p(x z) p(y z), although this holds by construction of p(x, y, z) and the derivation in Problem 3. Problem 8 Consider the probability that y is in a small interval [ȳ, ȳ + y] (with y > small), Pr (y [ȳ, ȳ + y]) = ȳ+ y ȳ p y (ỹ) dỹ, which approaches p y (ȳ) y in the limit as y. Let x := g(ȳ). For small y, we have by a Taylor series approximation, g(ȳ + y) g(ȳ) + dg dy (ȳ) y = x + x, x < }{{} =: x Since g(y) is decreasing, x <, see Figure. x x + x g(y) ȳ ȳ + y Figure : Decreasing function g(y) on a small interval [ȳ, ȳ + y]. 9

10 We are interested in the probability of x [ x + x, x]. For small x, this probability is equal to the probability that y [ȳ, ȳ + y] (since y [ȳ, ȳ + y] g(y) [g(ȳ), g(ȳ + y)] x [ x + x, x] for small y and x). Therefore, as y (and thus x ), Pr (y [ȳ, ȳ + y]) = p y (ȳ) y = Pr (x [ x + x, x]) = p x ( x) x p y (ȳ) y = p x ( x) x p x ( x) = p y(ȳ) x y As y (and thus x ), x p x ( x) = p y(ȳ) dg dy (ȳ) or p x(x) =. y dg dy p y(y) dg dy (y). (ȳ) and therefore Problem 9 From the change of variables formula, we have p(y) = p(x) dg dx (x) since g is continuously differentiable and strictly monotonic. Furthermore, y = g(x) dy = dg dx (x)dx. Let Y = g(x ), that is Y = {y x X : y = g(x)}. Then E [y] = Y ȳp y (ȳ) dȳ = X g( x) p x( x) dg dg dx ( x) dx ( x) d x = X g( x)p x ( x) d x. Problem Using the law of the unconscious statistician, we have for a discrete random variable x, E [ax + b] = x X and for a continuous random variable x, E [ax + b] = (ax + b)p(x) = a xp(x) +b p(x) = ae [x] + b x X x X }{{}}{{} E[x] (ax + b)p(x) dx = a xp(x) dx +b } {{ } E[x] p(x) dx = ae [x] + b. } {{ }

11 Problem Using the law of the unconscious statistician applied to the vector random variable PDF p(x, y) we have, [ ] x with y E [g(x)h(y)] = x,y = = g(x)h(y)p(x, y) dx dy g(x)h(y)p(x) p(y) dx dy (by independence of x, y) g(x)p(x) dx = E x [g(x)] E y [h(y)]. And similarly for discrete random variables. h(y)p(y) dy Problem 2 We will construct a counter-example. Let x be a discrete random variable that takes values -,, and with probabilities 4, 2, and 4, respectively. Let w be a discrete random variable taking the values -, with equal probability. Let x and w be independent. Let y be a discrete random variable defined by y = xw (it follows that y can take values -,, ). By construction, it is clear that y depends on x. We will now show that y and x are not independent, but E [xy] = E [x] E [y]. PDF p(x, y): x y p(x, y) x = (Prob 4 ) w = + (Prob 2 ) - impossible since w - 8 x = (Prob 4 ) w = (Prob 2 ) - impossible 2 w either ±, x = Marginalization p(x), p(y):. x p(x) y p(y)

12 It follows that p(x, y) = p(x) p(y), x, y does not hold. For example, p xy (, ) = 8 = p x ( ) p y (). Expected values: E [xy] = x xyp(x, y) y = ( ) ( ) 8 = + ( ) () + () () + () ( ) + () () E [x] = x xp(x) E [y] = = = So, for all x, y, E [xy] = E [x] E [y], but not p(x, y) = p(x) p(y). Hence, we have showed by counterexample that E [xy] = E [x] E [y] does not imply independence of x and y. Problem 3 From the definition of the probability density function for discrete random variables, p z ( z) = Pr (z = z). The probability that z takes on z is given by the sum of the probabilities of all possibilities of x and y such that z = x + y, that is all ȳ Y such that x = z ȳ and y = ȳ (or, alternatively, x X s.t. x = x and y = z x). That is, p z ( z) = Pr (z = z) = ȳ Y Pr (x = z ȳ)pr (y = ȳ) (by independence assumption) = ȳ Y p x ( z ȳ) p y (ȳ) Rewriting the last equation yields p z ( z) = ȳ Y p x ( z ȳ) p y (ȳ). Similarly, p z ( z) = x X p x ( x) p y ( z x) may be obtained. Problem 4 For t, we write F m (t) = Pr (m [, t]) = Pr (max{x,..., x n } [, t]) = Pr (x i [, t] i {,..., n}) = Pr (x [, t]) Pr (x 2 [, t]) Pr (x n [, t]) = t n. By substituting m for t, we obtain the desired result. 2

13 m Since F m ( m) = p m (τ) dτ, the PDF of m is obtained by differentiating, p m ( m) = df m dm ( m) = d d m ( mn ) = n m n. Problem 5 Let x = E [x]. Then Var [x] = (x x) 2 p(x) dx = = E [ x 2] 2 x 2 + x 2 = E [ x 2] x 2 Since Var [x], we have E [ x 2] = x 2 + Var [x] x 2, with equality if Var [x] =. x 2 p(x) dx 2 x xp(x) dx + x 2 } {{ } = x p(x) dx } {{ } = Problem 6 For fixed ȳ, Pr (x ȳ) = F x (ȳ). The probability that y [ȳ, ȳ + y] is Pr (y [ȳ, ȳ + y]) = ȳ+ y ȳ p y (ỹ) dỹ p y (ȳ) y for small y. Therefore, the probability that x ȳ and y [ȳ, ȳ + y] is Pr (x ȳ y [ȳ, ȳ + y]) = Pr (x ȳ) Pr (y [ȳ, ȳ + y]) F x (ȳ) p y (ȳ) y (3) by independence of x and y. Now, since we are interested in Pr (x y), i.e. in the probabilty of x being less than any y and not just a fixed ȳ, we can sum (3) over all possible intervals [ȳ i, ȳ i + y] Pr ( ȳ i such that x ȳ i ) = lim By letting y, we obtain Pr (x y) = F x (ȳ) p y (ȳ) dȳ. N N i= N F x (ȳ i ) p y (ȳ i ) y 3

14 Problem 7 First, we compute the mean of xy, E [xy] = E [x] E [y] = µ xµ y. x,y Hence, by using independence of x and y and the result of Problem 9, [ Var [xy] = E (xy) 2] (µ x µ y ) 2 x,y x,y = E [ x 2] E [ y 2] µ 2 xµ 2 y. Using E [ x 2] = σ 2 x + µ 2 x and E [ y 2] = σ 2 y + µ 2 y, we obtain the result Var x,y [xy] = ( σ 2 x + µ 2 x) ( σ 2 y + µ 2 y) µ 2 x µ 2 y = σ 2 xσ 2 y + µ 2 yσ 2 x + µ 2 xσ 2 y. Problem 8 a) The cumulative distribution is ˆF x ( x) = x λe λt dt = [ e λt] x t= = e λ x. Let u be a sample from a uniform distribution. According to the method (A3) presented in class, we obtain a sample x s by solving u = ˆF x (x s ) for x s, that is, u = e λxs e λxs = u λx s = ln ( u) x s = Notice the following properties: u x s, u x s +. ln( u). λ b) According to algorithm (A4) presented in class, we will use ˆp xy ( x, ȳ) = ˆp y x (ȳ x) ˆp x ( x) to obtain the sample x s from ˆp x and u, and the sample y s from ˆp y x (ȳ x s ) and u 2. We calculate x ˆp xy ( x, ỹ) dỹ = λ e λ x e λ 2 x x λ 2 e λ 2ỹ dỹ = λ e λ x, and, therefore, ˆp x ( x) = { λ e λ x x otherwise. 4

15 Furthermore, for x, ˆp y x (ȳ x) = ˆp xy( x, ȳ) ˆp x ( x) = { λ 2 e λ 2(ȳ x) x ȳ otherwise, with the corresponding CDF ˆF y x (ȳ x) = ȳ ȳ ˆp y x (t x) dt = λ 2 e λ2(t x) dt = e λ2(ȳ x). x Since the PDF ˆp x ( x) is identical to the PDF considered in part (a), we obtain the sample x s from u according to the solution from (a); that is, x s = ln( u ) λ. Notice that x s. We then obtain the sample y s by setting u 2 = ˆF y x (y s x s ) and solving for y s similarly to the solution in part (a): y s = ln( u 2) λ 2 + x s. Problem 9 For all x, ȳ {, } we have by marginalization, p x ( x) = p xy ( x, ȳ) = p xy ( x, ) + p xy ( x, ) = p xy ( x, ȳ) + p xy ( x, ȳ) p y (ȳ) = Therefore, ȳ {,} x {,} p xy ( x, ȳ) = p xy ( x, ȳ) + p xy ( x, ȳ). p x ( x) + p y (ȳ) = 2p xy ( x, ȳ) + p xy ( x, ȳ) + p xy ( x, ȳ) = 2p xy ( x, ȳ) + p xy ( x, ȳ) + p xy ( x, ȳ) + p xy ( x, ȳ) p xy ( x, ȳ) = p xy ( x, ȳ) + p xy ( x, ȳ) + p xy ( x, ȳ) + p xy ( x, ȳ) }{{} = p xy ( x, ȳ) = ȳ {,} x {,} p xy ( x, ȳ) + p xy ( x, ȳ) = + p xy ( x, ȳ) p xy ( x, ȳ) + p xy ( x, ȳ) = p xy ( x, ȳ) p x ( x) + p y (ȳ). 5

16 Problem 2 E [x] = = a a a xp(x) dx = xp(x) dx a xp(x) dx + } {{ } a ap(x) dx p(x) dx = a Pr (x a) a xp(x) dx Problem 2 This inequality can be obtained using Markov s inequality: (x x) 2 is a nonnegative random variable; apply Markov s inequality with a = k 2 (k ): Pr ( (x x) 2 k 2) E [ (x x) 2] k 2 = Var [x] k 2 = σ2 k 2 Since (x x) 2 k 2 x x k, the result follows: Pr ( x x k) σ2 k 2. Problem 22 Consider the random variable x +x 2 + +x n n. It has the mean [ ] x + x x n E = n n (E [x ] + + E [x n ]) = n (nµ) = µ and variance [ ] [ (x x + x x n + x x n Var = E n n = [ (E n 2 (x µ) (x n µ) 2 + cross terms The expected value of the cross terms is zero, for example, Hence, ) ] 2 µ ]). E [(x i µ) (x j µ)] = E [x i µ] E [x j µ] for i j, by independence =. [ ] x + x x n Var = n n 2 (Var [x ] +... Var [x n ]) = nσ2 n 2 = σ2 n 6

17 Applying Chebyshev s inequality with ɛ > gives ( ) x + x x n Pr µ n ɛ σ2 nɛ 2 n ( ) x + x x n Pr µ n > ɛ n Remark: Changing to > is valid here, since ( ) ( ) x + + x n Pr µ n ɛ x + + x n = Pr µ n > ɛ ( ) x + + x n + Pr µ n = ɛ. }{{} = since x + +xn n is a CRV Problem 23 First, we rewrite Pr (5 < x < 5) = Pr ( x < 5) = Pr ( x 5). Using Chebyshev s inequality, with x =, σ 2 = 5, k = 5, yields Pr ( x 5) = 5 25 = 3 5. Therefore, we can say that Pr (5 < x < 5) 3 5 = 2 5. Problem 24 For s, t, Pr (x > s + t x > t) Pr (x > s + t) Pr (x > s + t x > t) = = Pr (x > t) Pr (x > t) [ s+t λe λ x d x e λ x ] = t λe λ x d x = s+t [ e λ x ] = e λ(s+t) t e λt = e λs = s = Pr (x > s). λe λ x d x 7