Continuous Random Variables

Transcription

1 Probability 2 - Notes 7 Continuous Random Variables Definition. A random variable X is said to be a continuous random variable if there is a function f X (x) (the probability density function or p.d.f.) mapping the real line R into [0, ) such that for any open interval (a,b), P(X (a,b)) = P(a < X < b) = b a f X (x)dx. From the axioms of probability this gives: (i) f X (x)dx =. (ii) The cumulative distribution function F X (x) = P(X x) = x f X (u)du. F X (x) is a monotone increasing function of x with F X () = 0 and F X ( ) =. (iii) P(X = x) = 0 for all real x. From calculus, df X (x) dx c.d.f. is differentiable. for all points for which the p.d.f. is continuous and hence the Expectations, Moments and the Moment Generating Functions E[g(X)] = g(x) f X (x)dx The raw moments are moments about the origin. The r th raw moment is µ r = E[X r ]. Note that µ is just the mean µ. The moment generating function (m.g.f.) M X (t) = E[e tx ]. For a discrete random variable M X (t) = G X (e t ). For a continuous random variable M X (t) = e tx f X (x)dx. Properties of the M.G.F. (i) If you expand M X (t) in a power series in t you obtain M X (t) = r=0 µ rt r r!. So the m.g.f. generates the raw moments. (ii) µ r = E[X r ] = M (r) X (0), where M(r) X (t) denotes the rth derivative of M X (t) with respect to t. (iii) The m.g.f. determines the distribution. Other properties (similar to those for the p.g.f.) will be considered later once we have looked at joint distributions.

2 Standard Continuous Distributions Uniform Distribution. All intervals (within the support of the p.d.f.) of equal length have equal probability of occurrence. Arises in simulation. Simulated values u j } from a uniform distribution on (0,) can be transformed to give simulated values x j } of a continuous r.v. X with c.d.f. F by taking x j = F (u j ). X U(a,b) if (b a) if a < x < b E[X] = a+b 2 and Var(X) = (b a)2 2. M X (t) = ebt e at t(b a). This exists for all real t. Exponential Distribution. Used for the time till the first event if events occur randomly and independently in time at constant rate. Used as a survival distribution for an item which remains as good as new during its lifetime. X Exp(θ) if θe θx if 0 < x < E[X] = θ and Var(X) = θ 2. M X (t) = ( t θ). This exists for t < θ. Gamma Distribution. Exponential is special case. Used as a survival distribution. When α = n, gives the time until the n th event when events occur randomly and independently in time. X Gamma(θ,α) if θ α x α e θx Γ(α) if 0 < x < The Gamma function is defined for α > 0 by Γ(α) = 0 x α e x dx. It is then simple to show that the p.d.f. integrates to one by making a simple change of variable (y = θx) in the integral. It is easily shown using integration by parts that Γ(α + ) = αγ(α). Therefore when n is a positive integer Γ(n) = (n )!. E[X] = α θ and Var(X) = α θ 2. 2

3 M X (t) = ( t θ) α. This exists for t < θ. Note: The Chi-squared distribution (X χ 2 n) is just the gamma distribution with θ = /2 and α = n/2. This is an important distribution in normal sampling theory. Normal Distribution. Important in statistical modelling where normal error models are commonly used. It also serves as a large sample approximation to the distribution of efficient estimators in statistics. X N(µ,σ 2 ) if (x µ)2 e 2σ 2 σ 2 To show that the p.d.f. integrates to by a simple change of variable in the integral to z = (x µ)/σ we just need to show that e z2 /2 =. We show this at the end of Notes 5. E[X] = µ and Var(X) = σ 2. M X (t) = e µt+ σ2 t 2 2. This exists for all t. Example deriving the m.g.f. and finding moments X N(µ,σ 2 ). M X (t) = e tx σ 2 e (x µ)2 /(2σ 2) dx In the integral make the change of variable to y = (x µ)/σ. Then M X (t) = e y2 /2+t(µ+σy) = e µt+σ2 t 2 /2 e (y σt)2 /2 = e µt+σ2 t 2 /2 Finding E[X] and E[X 2 ]. Differentiating gives M X (t) = (µ + σ2 t)e µt+σ2 t 2 /2 and M (2) X (t) = (σ2 )e µt+σ2 t 2 /2 + (µ + σ 2 t) 2 e µt+σ2 t 2 /2 Therefore E[X] = M X (0) = µ and E[X 2 ] = M (2) X (t) = σ2 + µ 2. 3

4 Transformations of random variables. Theorem. Let the interval A be the support of the p.d.f. f X (x). If g is a : continuous map from A to an interval B with differentiable inverse, then the r.v. Y = g(x) has p.d.f. f Y (y) = f X (g (y)) Proof.This is easily shown using equivalent events. The function g(x) will either be (a) strictly monotone increasing; or (b) strictly monotone decreasing. We consider each case separately. Case (a) F Y (y) = P(Y y) = P(g(X) y) = P(X g (y)) = F X (g (y)) Differentiating and noting that dg (y) > 0 gives f Y (y) = f X (g (y)) dg (y) = f X (g (y)) Case (b) F Y (y) = P(Y y) = P(g(X) y) = P(X g (y)) = F X (g (y)) Differentiating and noting that dg (y) < 0 gives f Y (y) = f X (g (y)) dg (y) = f X (g (y)). Example X N(µ,σ 2 ). Let Y = X µ σ. Then g (y) = µ + σy. Therefore dg (y) = σ. Hence f Y (y) = f X (g (y)) = σ 2 e y2 /2 σ = e y2 /2 The support for the p.d.f. X, (, ), is mapped onto (, ) so this is the support of the p.d.f. for Y. Therefore Y N(0,). Transformations which are not :. You can still find the c.d.f. for the transformed variable by writing F Y (y) as an equivalent event in terms of X. Example. X N(0,) and Y = X 2. The support for the p.d.f. of Y is [0, ). For y > 0, F Y (y) = P(X 2 y) = P( y X y) = F X ( y) F X ( y) 4

5 Differentiating with respect to y gives, for y > 0, f Y (y) = f X ( y) 2 y f X( y) 2 y = y /2 e y/2 2 /2 π This is just the p.d.f. for a χ 2. Note that this implies that Γ(/2) = π because the constant in the p.d.f. is determined by the function of y and the range (suppport of the p.d.f.) since the p.d.f. integrates to one. Note for the normal p.d.f. Let A = e z2 /2 dz. Note that A > 0. Then A 2 = e (x2 +y 2 )/2 dx Making the change to polar co-ordinates (Calculus 2) gives A 2 = 0 0 e r2 /2 rdrdθ = Hence A =. 5