Continuous Random Variables

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2 Normal Approximation to the Binomial

3 Definition X is continuous if there exists a 0 function f (x) such that P(X A) = f (x)dx. This function is called the probability density function, or pdf. Properties: f (x)dx = 1 (= P( < X < )). P(a X b) = P(a < X < b) = b P(X = a) = a a f (x)dx = 0. Read Example 1a, p A a f (x)dx.

4 Example X = life time of an indicator light, has pdf f (x) = 100 x I (x > 100) 1. Find P(X < 150) 2. What is the probability that two of 5 such indicator lights will have to be replaced within the first 150 hours? Definition F (x) = P(X x) = x distribution function. f (t)dt is called the cumulative d By the Fundamental Theorem of Calculus, dx F (x) = f (x)

5 Example X f X (x). Find the pdf of Y = 2X. µ X = E(X ) = xf X (x)dx (Definition) E(g(X )) = g(x)f (x)dx (To be shown see next page) E(ag 1 (X ) + bg 2 (X )) = ae(g 1 (X )) + be(g 2 (X )) Example X f X (x) = I (0 x 1). Find E(e X ). Solution: a) Apply above formula with g(x) = e x b) Find the pdf, f Y (y), of Y = e X and use the definition of expected value, i.e. E(Y ) = yf Y (y)dy.

6 Lemma If Y is a 0 r.v., then E(Y ) = 0 P(Y > y)dy Proposition E(g(X )) = g(x)f (x)dx. Proof: Assume first that g(x) 0, for all x. Then, E(g(X )) = P(g(X ) > y)dy = = 0 g(x) 0 dyf (x)dx = 0 x:g(x)>y g(x)f (x)dx f (x)dxdy For general g write g(x) = g(x)i (g(x) 0) + g(x)i (g(x) < 0) = g + (x) g (x), and use E(g(X )) = E(g + (X )) E(g (X ))

7 σ 2 X = Var(X ) = E(X µ X ) 2 = E(X 2 ) µ 2 X. Var(aX + b) = a 2 Var(X ) Example (The Uniform in (0,1) Random Variable) X U(0, 1) if f X (x) = I (0 < x < 1). Show F X (x) = 0I (x 0) + xi (0 < x 1) + I (x > 1), µ X = 0.5, σ 2 X = 1 12 Example (The Uniform in (a,b) Random Variable) X U(0, 1) if f X (x) = 1 b ai (a < x < b). Derive expressions for F X, µ X and σx 2

8 Normal Approximation to the Binomial X N(µ, σ 2 ) if f X (x) = 1 2πσ exp{ (x µ)2 2σ 2 } If µ = 0, σ 2 = 1, X is said to have the standard normal distribution. A standard normal random variable is typically denoted by Z. Showing that f X integrates to one is not simple. In fact, you either know how to do this integral or you spend a lot of time going nowhere. See page 199 of the book. If X N(µ, σ 2 ) then Y = a + bx N(a + bµ, b 2 σ 2 )

9 Normal Approximation to the Binomial Corollary 1. If Z N(0, 1), then X = µ + σz N(µ, σ 2 ). 2. If X N(µ, σ 2 ), then Z = X µ σ 3. If X N(µ, σ 2 ), then x α = µ + σz α, N(0, 1). where x α and z α denote the percentiles of X and Z.

10 Normal Approximation to the Binomial Finding Probabilities via the Standard Normal Table In Table A.3, z-values are identified from the left column, up to the first decimal, and the top row, for the second decimal. Thus, 1 is identified by 1.0 in the left column and 0.00 in the top row. Example (The % Property.) Let Z N(0, 1). Then 1. P( 1 < Z < 1) = Φ(1) Φ( 1) = = P( 2 < Z < 2) = Φ(2) Φ( 2) = = P( 3 < Z < 3) = Φ(3) Φ( 3) = =.9974.

11 Normal Approximation to the Binomial Finding Probabilities via the Standard Normal Table Example Let X N(1.25, ). Find a) P(1 X 1.75), and b) P(X > 2). Solution. Use Z = X 1.25 N(0, 1) to express these 0.46 probabilities in terms of Z. Thus, ( a) P(1 X 1.75) = P X 1.25 ) = P(.54 < Z < 1.09) = Φ(1.09) Φ(.54) = ( b) P(X > 2) = P Z > ) = 1 Φ(1.63) =

12 Normal Approximation to the Binomial Finding Percentiles via the Standard Normal Table To find z α, one first locates 1 α in the body of Table A.3 and then reads z α from the margins. If the exact value of 1 α does not exist in the main body of the table, then an approximation is used as described in the following. Example Find z 0.05, the 95th percentile of Z. Solution. 1 α = 0.95 does not exist in the body of the table. The entry that is closest to, but larger than 0.95 (i.e ), corresponds to The entry that is closest to, but smaller than 0.95 (which is ), corresponds to We approximate z by averaging these two z-values: z.05 =

13 Normal Approximation to the Binomial Finding Percentiles via the Standard Normal Table Example Let X denote the weight of a randomly chosen frozen yogurt cup. Suppose X N(8,.46 2 ). Find the value c that separates the upper 5% of weight values from the lower 95%. Solution. This is another way of asking for the 95-th percentile, x.05, of X. Using the formula x α = µ + σz α, we have x.05 = z.05 = 8 + (.46)(1.645) = 8.76.

14 Normal Approximation to the Binomial The Basic Result If X Bin(n, p), then the DeMoivre-Laplace limit theorem states that X N(np, np(1 p)), for n large enough. The approximation is quite good for values of n such that np(1 p) 10 and is often used if np 5 and n(1 p) 5. If p = X /n, the DeMoivre-Laplace limit theorem also yields ( ) p p(1 p) N p,, for n large enough as above. n

15 Normal Approximation to the Binomial The Continuity Correction Due to the discreteness of the binomial distribution, the normal approximation is improved by the so-called continuity correction: P(X x) = P(X x + 0.5) ( ) x np = P(Y x + 0.5) = Φ, np(1 p) where Y N(np, np(1 p)), i.e. is a normal r.v. with mean and variance equal to the mean and variance of the Binomial r.v. X. The normal approximation works in situations where the Poisson approximation does not work. For example p does not have to be 0.01.

16 Normal Approximation to the Binomial Example Suppose that 60% of all drivers in a certain state wear seat belt always. A random sample of 500 drivers is selected. Find the (approximate) probability that the number of those wearing seat belt always is between 270 and 320 (inclusive). Solution: Let X denote the number of drivers that always wear seat belt. Then, X Bin(500, 0.6). Since np 5 and n(1 p) 5, P(270 X 320) = P(X 320) P(X 269) ( ) ( ) = Φ Φ = Φ(1.826) Φ( 2.83) = =

17 Normal Approximation to the Binomial Example (Continued) Using the continuity correction, we have P(270 X 320) = P(X 320) P(X 269) ( ) ( ) = Φ Φ = Φ(1.87) Φ( 2.78) = =

18 X Exp(λ) if f X (x) = λe λx I (x > 0), for some λ > 0. F (x) = P(X x) = 1 e λx E(X n ) = n λ E(X n 1 ), so that E(X ) = 1 λ, Var(X ) = 1 λ 2 Example Suppose that the number of miles a car can run before its battery wears out is exponentially distributed with an average value of 10,000 miles. A person decides to take a 5,000 mile trip having just changed the battery. What is the probability that the trip will be completed without having to replace the battery? Solution: P(X > 5) = e 5/10 =

19 The Memoryless Property of the Exponential RV If X Exp(λ) then for t > s we have Example P(X > t X > s) = P(X > t s) Suppose that the number of miles a car can run before its battery wears out is exponentially distributed with an average value of 10,000 miles. A person decides to take a 5,000 mile. What is the probability that the trip will be completed without having to replace the battery? Solution: By the memoryless property, P(X > 5) = e 5/10 =

20 The Poisson-Exponential Relationship Proposition Let X (t) be a Poisson process with parameter λ, and let T be the time until the first occurrence. Then T Exp(λ)

21 Example a) If X U(0, 1), find the distribution of Y = X 2. b) If X U( 1, 1), find the distribution of Y = X 2. c) If X U(0, 1), find the distribution of Y = log X. Theorem Let X be continuous with pdf f X, and let g(x) be strictly monotonic and differentiable function. Then Y = g(x ) has pdf f Y (y) = f X (g 1 (y)) d dy g 1 (y) for y in the range of the function g, and zero otherwise.

22 Example 1. (The Probability Transformation) Let X be continuous with cumulative distribution function F X. Then, if g = F X, Y = g(x ) U(0, 1). 2. (The Quantile Transformation) Let X U(0, 1) and F be a cumulative distribution function of a continuous random variable. Then, if g = F 1, Y = g(x ) has F Y = F.