Generating Random Variates 2 (Chapter 8, Law)

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1 B. Maddah ENMG 6 Simulation /5/08 Generating Random Variates (Chapter 8, Law) Generating random variates from U(a, b) Recall that a random X which is uniformly distributed on interval [a, b], X ~ U(a, b), has the distribution function: 0, F( x) ( x a) /( b a),, if x a if a x b if x b To utilize the inverse transform method, set u = F(x) for a x b, and solve for x, u = ( x a) /( b a) x = a + u( b a). Then, the algorithm for generating X ~ U(a, b) is as follows:. Generate U ~ U(0,). Set X = a + U(b a). Generating random variates from triang(a, b, m) A r.v. X with a triangular distribution, X ~ triang(a, b, m), has the following distribution function: a m b ( x a) /( b a)( m a), x) ( b x) /( b a)( b m), F X ( if if a x m m x b

2 Fact. If X ~ triang(a, b, m), then X = a +(b a)y, where Y ~ triang (0,, m ), where m = (m a) /(b a). Proof. Note that P{ Y Then, y / m, < y} ( y) /( m ), [( ) /( )], if 0 if 0 y m if m y P{ a+ ( b a) Y < x} = P{ Y < ( x a)/( b a)} [( x a) /( b a)] x a, if 0 m m ( b a) [ ( x a) /( b a)] x a, if m m ( b a) x a b a x a m a ( m a)/( b a) ( b a) ( b a) [ ( x a) /( b a) ] m a x a, if ( m a)/( b a) ( b a) ( b a) ( x a) /( b a)( m a), ( b x) /( b a)( b m), = P{X < x}. if a x m if m x b To generate Y, using the inverse transform method, solve u = FY ( y) = y / m, if 0 y m ( y) /( m), if m y.

3 y m u, ( u)( m ), if 0 if m m u m ( u)( m ) y mu, ( u)( m ), if 0 u m if m u Finally, this is the algorithm for generating triang(a,b,m).. Set m = (m a) /(b a).. Generate U ~ U(0,). 3. If U < m, set = m U. Otherwise, set Y Y = ( U)( m ). 4. Set X = a +(b a)y. Generating random variates from N(μ, σ) Recall that if X ~ N(μ, σ), then X = μ +σz, where Z ~ N(0,). Once Z is generated, X can be generated as follows:. Generate Z ~ N(0,).. Set X = μ +σz. In the following we present two algorithms for generating Z. The first algorithm is based on the central limit, which implies that if U i are iid U(0,) then U i 6 N(0, ). i= 3

4 This (approximate) algorithm works as follows. Generate U, U,..., U ~ U(0,). Set Z = U i 6. i= The second (Box-Muller) algorithm is based on the following fact. Fact. Let Z and Z be independent N(0,) rvs. Let R, = Z + Z θ = tg ( Z / Z ). Then, R is an exponential rv with mean (R ~ exp(/)), and θ ~ U(0, π). Z θ R Noting that Z = R cos(θ), and Z = R sin(θ), pairs of Z can be generated as follows.. Generate U, U ~ U(0,). Set Z = ln( U )] / cos(π ), [ U Z Z [ U = ln( U )] / sin(π ). 4

5 Generating random variates from LN(μ, σ) A rv X is said to have a lognormal distribution, X ~LN(μ, σ), if Y = ln(x) ~ N(μ, σ). The, X can be generated as follows.. Generate Y ~ N(μ, σ).. Set X = e Y. Generating random variates from gamma(α, β) A rv X with a gamma distribution, X ~ gamma(α, β), has the following density and distribution function: α α x/ β x α α t/ β β x e β t e f X( x) =, x> 0, FX( x) = dt, Γ( α) Γ( α) 0 where z t Γ ( α) = t e dt. 0 Since F X (x) generally has no closed form, the inverse transform method cannot be applied except numerically. When α is an integer, then X is an Erlang r.v. representing the sum of α exponential random variables each having a mean β. I.e., X ~ α-erlang(/β). In this case the convolution method for generating Erlang can be used. If α is not an integer or α is a large integer, the acceptance-rejection method can be used. 5

6 Since X ~ gamma(α, β) can be written as X = βy, where Y~ gamma(α, ). Then, the focus is on generating Y. To generate Y via acceptance-rejection, consider two cases: α < and α >. (If α =, then Y ~ exp()). If α <, then the following majorizing function is used: α x, if 0< x Γ ( α) tx ( ) x e if x > Γ( α) 3 f X x, t x, x If α >, then the following majorizing function is used: λ λμx tx ( ) = c, λ ( μ + x ) α α λ where λ = α, c= 4 α e /( λγ ( α)), μ = α. 6

7 0.4 f X ( x, ) tx (, ) x Generating from the density function r(x) based on t(x) in both cases can be done with the inverse-transform method. (see Law, pp , for details). Generating random variates from beta(α,α ) If X ~ beta(α,α ) (see Law, p. 0 for description), then X can be written as X = Y /(Y +Y ), where Y ~ gamma(α, ) and Y ~ gamma(α, ). Then, X can be generated as follows:. Generate Y ~ gamma(α, ) and Y ~ gamma(α, ).. Set X = Y /(Y +Y ). Generating random variates from a discrete Uniform rv An integer-valued rv, X, is said to be discrete uniform if it is equally likely to take on any integer between integers i and j, i < j. 7

8 Note that the pmf of X, is P{X = k} = /(j i+), k = i, i+,, j. Applying the inverse transform method gives the following algorithm:. Generate U ~ U(0,).. Set X = i+ ( j i+ ) U, where x is the largest integer x (e.g..3 = ). Generating random variates from a Bernoulli rv Recall that a Bernoulli rv takes on values 0 or with probabilities p and p <. Applying the inverse transform method gives the following algorithm:. Generate U ~ U(0,).. If U p, set X =. Otherwise, set X = 0. Generating random variates from a Binomial rv Recall that if X has a binomial distribution with parameters n and p, X n = Yi, where Y i ~ Bernoulli with parameter p. i= Then, X can be generated as follows.. Generate Y, Y,, Y n ~ Bernoulli(p). Set X n = Yi. i= 8

9 Generating from a Geometric rv The distribution function of the Geometric rv is F () i = P{ X i} = ( p) i X Then, the inverse-transform method can be applied as follows to generate X. Let q = p.. Generate U ~ U(0,).. If q i < U q i, set X = i. Note that in step, X is the smallest i such that U F(i) = q i, which implies that X is the smallest i such that q i < U. Solving q i = U for i implies that i =ln( U)/ln(q). Then, since q i is decreasing in i, the smallest value of i such that q i < U is X = ln( U) / ln( q), where x is the smallest integer x. This leads to the following algorithm for generating X.. Generate U ~ U(0,).. Set X = ln( U) / ln( q). 9

10 Generating from a Poisson rv Recall that the pmf for the Poisson rv with parameter λ is i λ λ px () i = P{ X = i} = e, i= 0,, K i! λ Note that px( i+ ) = px( i) and i + i FX() i = px( j). j= 0 Then, since F X (i) is increasing in i, the smallest value of i for which U F X (i), can be found by first comparing U and F X (0) = p X (0) = e -λ, and then, if U > p X (0), comparing U with F X () = p X (0) + λp X (0), and so on. The algorithm works as follows.. Set p = e λ, F = p, i = 0.. Generate U ~ U(0,). 3. If U < F, set X = i and stop. 4. Set p = pλ/(i+), F = F + p, i = i+ 5. Go to Step 3. 0