AMS 216 Stochastic Differential Equations Lecture 02 Copyright by Hongyun Wang, UCSC ( ( )) 2 = E X 2 ( ( )) 2

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1 AMS 216 Stochastic Differetial Equatios Lecture 02 Copyright by Hogyu Wag, UCSC Review of probability theory (Cotiued) Variace: var X We obtai: = E X E( X ) 2 = E( X 2 ) 2E ( X )E X var( X ) = E X 2 Stadard deviatio: std( X ) = var( X ) E X ( ) 2 = E X 2 2X E( X )+ E X ( ) 2 = E X 2 + E X ( ) 2 ( ) 2 E X Properties of E(X) ad var(x) i) E(aX + by) = ae(x) + be(y) This is valid for all X ad Y. I particular, X ad Y do ot eed to be idepedet. ii) If X ad Y are idepedet, the we have E(X Y) = E(X) E(Y) Proof: = xyρ X,Y E XY ( x, y)dxdy = xyρ X ( x)ρ Y ( y)dxdy ( yρ Y ( y)dy) = E X = xρ X ( x)dx E Y Cautio: E(X Y) = E(X) E(Y) does ot imply that X ad Y are idepedet. Example: - 1 -

2 AMS 216 Stochastic Differetial Equatios ( X,Y ) = ( 0,1), Pr = 0.25 ( 0, 1), Pr = 0.25 ( 1,0), Pr = 0.25 ( 1,0), Pr = 0.25 E(X) = 0, E(Y) = 0, E(X Y) = 0 But X 2 + Y 2 = 1 iii) If X ad Y are idepedet, the we have Proof: var(x + Y) = var(x) + var(y) var(x + Y) = E((X + Y) 2 ) (E (X + Y)) 2 = Complete the proof i your homework. Examples of distributios: 1) Beroulli distributio 1, Pr = p X = 0, Pr = 1 p Example: Flip a coi 1: head, success 0: tail, failure Rage = {0, 1} Expected value ad variace: E(X) = p, E(X 2 ) = p Var(X) = E(X 2 ) (E(X)) 2 = p(1 p) 2) Biomial distributio N = sum of idepedet Beroulli radom variables N = i=1 X i PMF (probability mass fuctio): - 2 -

3 AMS 216 Stochastic Differetial Equatios Pr( N = k) = C(,k)p k ( 1 p) k Example: # of heads i flips Rage = {0, 1, 2,, } Expected value ad variace: = E( X 1 + X 2 +!+ X ) = p = var( X 1 + X 2 +!+ X ) = p( 1 p) E N var N 3) Expoetial distributio PDF (probability desity fuctio): ρ T λexp( λt ), t 0 ( t ) = 0, t < 0 Example: (Escape problem) T = time util escape from a deep potetial well by thermal fluctuatios Mathematical defiitio: T = time util occurrece of a evet i a memoryless system We derive the PDF based o the memoryless property. memoryless meas ( t T >t 0 ) = Pr( T t ) Pr T t 0 Cosider the complemetary cumulative distributio fuctio (CCDF) G( t ) Pr( T >t ) = ρ t G(0) = Pr(T > 0) = 1 t d t We re-write the memoryless property i terms of G(t). Pr( ( T t 0 ) t AND T >t 0 ) = Pr T t Pr( T >t 0 ) ==> Pr( t 0 <T t 0 +t ) = Pr( T t )Pr( T >t 0 ) ==> G( t 0 ) G t 0 +t G t 0 = 1 G( t ) Replace t with Δt, divide by Δt, ad take the limit as Δt 0, we get - 3 -

4 G( t 0 ) G t 0 + Δt Δt ==> G t 0 = G ( 0) λ AMS 216 Stochastic Differetial Equatios!" # $# G t 0 = G ( 0 ) G Δt Δt We obtai a iitial value problem (IVP) = λg( t 0 ) = 1 G t 0 G 0 The solutio is G( t ) = exp( λt ). Differetiate G(t), we obtai ρ( t ) = d dt G ( t ) = Expected value ad variace: CDF: E( T ) = 1 λ, var ( T ) = 1 λ 2 F T ( t ) = Pr T t λexp( λt ), t 0 0, t < 0 G( t 0 ) = 1 exp( λt ) for t 0 4) Normal distributio PDF: ρ X ( x) = 2 1 exp x µ 2πσ 2 2σ 2 Notatio: X ~ N(μ, σ 2 ) Example: (Cetral Limit Theorem) Suppose {X1, X2,, XM} are i.i.d. (idepedetly ad idetically distributed). Whe M is large, approximately X = X j has a ormal distributio. Rage = (, + ) M - 4 -

5 AMS 216 Stochastic Differetial Equatios Expected value ad variace: E( X ) = xρ( x)dx = µ var X = E ( X µ ) 2 = ( x µ ) 2 ρ( x)dx = σ 2 CDF of ormal distributio: F X ( x) = Pr X x x = ρ X x dx = x 2 1 exp x µ 2πσ 2 2σ 2 d x Chage of variables: s = x µ, d x = 2σ 2 2σ 2 ds F X ( x) = x µ 2σ 2 exp s2 ds = 1π π The error fuctio: erf ( z) 1 π z exp( s 2 )ds = 2 π z z 0 x µ 2σ 2 0 exp( s2)ds With the error fuctio, we write the CDF as F X Example: ( x) = erf x µ 2σ 2 exp( s2)ds = erf µ + ησ µ Pr X µ + ησ 2σ 2 = 1 η 1+ erf 2 2 Properties of erf(z): i) erf(0) = 0 ii) erf(+ ) = 1 iii) erf( z) = erf(z) Shape of PDF: (Draw the bell-shaped PDF, showig the ceter μ, ad show the iterval [μ ησ, μ + ησ] ) - 5 -

6 AMS 216 Stochastic Differetial Equatios We like to fid η such that = % Pr X µ ησ We express this probability i terms of CDF. = Pr µ ησ X µ + ησ Pr X µ ησ = F X ( µ + ησ) F X µ ησ =! = erf η 2 We set erf η 2 = 0.95 ==> η= erfiv ( 0.95) 2 = 1.96 We obtai = 95% Pr X µ 1.96σ Similarly, we ca obtai = 99% Pr X µ σ Cofidece iterval: Suppose we are give a data set of idepedet samples of X ~ N(μ, σ 2 ). {Xj, j = 1, 2,, } Questio: How to estimate μ from data? ˆµ = 1 X j Questio: How to estimate the ucertaity i μ? μ is a radom variable. = E 1 E ˆµ var ˆµ X j = var 1 = 1 E X +!+ X 1 X j = µ = 1 var X +!+ X 2 1 = σ2 (Here we used the idepedece of {Xj}) Theorem: - 6 -

7 AMS 216 Stochastic Differetial Equatios Sum of idepedet ormal radom variable ormal radom variable This theorem will be proved later i the discussio of characteristic fuctios. It follows that μ is ormal. ˆµ ~ N µ, σ2 The iterval cotaiig 95% probability of μ is described by Pr ˆµ µ 1.96 σ = 95% Case 1: Suppose we kow the value of σ. ˆµ 1.96 σ, ˆµ σ is called the 95% cofidece iterval. Example: We are give a data set of 100 idepedet samples of X ~ N(μ, σ 2 ). {3.0811, , , } We are give σ = 1.3 μ is estimated as ˆµ = 1 X j = σ = The 95% cofidece iterval is (0.2202, ) Questio: What is the meaig of this cofidece iterval? The data set is give, fixed. μ is fixed, although ukow. What is the meaig of (0.2202, )? Two key compoets i iterpretig the cofidece iterval: i) The cofidece iterval is a algorithm/fuctio that maps data to iterval ˆµ L ({ X j }), ˆµ H ({ X j }) ii) The framework of repeated experimets. Draw a data set of idepedet samples of X ~ N(μ (True), σ 2 )

8 AMS 216 Stochastic Differetial Equatios Repeat this M times (M is large). The meaig of cofidece iterval is Pr ˆµ L ({ X j })!#" # $ < µ ( True )!" # $# < ˆµ H ({ X j }) = 0.95!#" # $ Fixed Radom variable Radom variable I other words, suppose we go over M data sets ad estimate the cofidece iterval for each data set. For 95% of data sets, the cofidece iterval cotais μ (True). A similar example: I have a algorithm for estimatig a iterval for a dog s age. Suppose we go over M dogs. For 95% of dogs, the dog s true age falls i the estimated iterval. I summary, the two key compoets are i) a algorithm, ii) repeated experimets. Case 2: σ is ukow Recall the defiitio of stadard deviatio. σ = var X = E ( X µ ) 2 We ca estimate σ as ˆσ = 2 1 X j ˆµ, ˆµ = 1 X j A approximate 95% cofidece iterval is ˆµ 1.96 ˆσ, ˆµ ˆσ Correspodigly, the oe calculated usig the exact value of σ (case 1 above) is called the exact 95% cofidece iterval ˆµ 1.96 σ, ˆµ σ - 8 -