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Example Suppose X ~ Uniform(2, 4). Let . Find .

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1 Example Suppose X ~ Uniform(2, 4). Let . Find .
What if X ~ Uniform(-4, 4)? STA286 week 7

2 Functions of Random variables
In some case we would like to find the distribution of Y = h(X) when the distribution of X is known. Discrete case Examples 1. Let Y = aX + b , a ≠ 0 2. Let STA286 week 7

3 Continuous case – Examples
1. Suppose X ~ Uniform(0, 1). Let , then the cdf of Y can be found as follows The density of Y is then given by 2. Let X have the exponential distribution with parameter λ. Find the density for 3. Suppose X is a random variable with density Check if this is a valid density and find the density of STA286 week 7

4 Question Can we formulate a general rule for densities so that we don’t have to look at cdf? Answer: sometimes … Suppose Y = h(X) then and but need h to be monotone on region where density for X is non-zero. STA286 week 7

5 Check with previous examples: 1. X ~ Uniform(0, 1) and
2. X ~ Exponential(λ). Let 3. X is a random variable with density and STA286 week 7

6 Theorem If X is a continuous random variable with density fX(x) and h is strictly increasing and differentiable function form R  R then Y = h(X) has density for Proof: STA286 week 7

7 Theorem If X is a continuous random variable with density fX(x) and h is strictly decreasing and differentiable function form R  R then Y = h(X) has density for Proof: STA286 week 7

8 Summary If Y = h(X) and h is monotone then Example X has a density
Let Compute the density of Y. STA286 week 7

9 Exercise Find the density of X = Z2 where Z ~ N(0,1).
This is the Chi-Square density with parameter 1. Notation: STA286 week 7

10 Change-of-Variable for Joint Distributions
Theorem Let X and Y be jointly continuous random variables with joint density function fX,Y(x,y) and let DXY = {(x,y): fX,Y(x,y) >0}. If the mapping T given by T(x,y) = (u(x,y),v(x,y)) maps DXY onto DUV. Then U, V are jointly continuous random variable with joint density function given by where J(u,v) is the Jacobian of T-1 given by assuming derivatives exists and are continuous at all points in DUV . STA286 week 7

11 Example Let X, Y have joint density function given by
Find the density function of STA286 week 7

12 Examples Suppose X~ Poisson(λ1) independent of Y~ Poisson(λ2). Find the distribution of X+Y. X, Y independent each having Exponential distribution with mean 1/λ. Find the density for W=X+Y. If X, Y independent standard normal random variables, find the density of W=X+Y. STA286 week 7

13 Density of Quotient Suppose X, Y are independent continuous random variables and we are interested in the density of Can define the following transformation The inverse transformation is x = w, y = wz. The Jacobian of the inverse transformation is given by Apply 2-D change-of-variable theorem for densities to get The density for Z is then given by STA286 week 7

14 Example Suppose X, Y are independent N(0,1). The density of is
STA286 week 7

15 Moments The kth moment of a distribution is E(Xk). It is given by…
We are usually interested in 1st and 2nd moments (sometimes in 3rd and 4th) week 5

16 Moment Generating Functions
The moment generating function of a random variable X is mX(t) exists if mX(t) < ∞ for |t| < t0 >0 If X is discrete If X is continuous STA286 week 7

17 Examples X ~ Exponential(λ). The mgf of X is
X ~ Uniform(0,1). The mgf of X is STA286 week 7

18 Generating Moments from MGFs
Theorem Let X be any random variable. If mX(t) < ∞ for |t| < t0 for some t0 > 0. Then mX(0) = 1 etc. In general, Where is the kth derivative of mX with respect to t. STA286 week 7

19 Examples Suppose X ~ Exponential(λ). Find the mean and variance of X using its moment generating function. Suppose X ~ N(0,1). Find the mean and variance of X using its moment generating function. Suppose X ~ Binomial(n, p). Find the mean and variance of X using its moment generating function. STA286 week 7

20 Properties of Moment Generating Functions
mX(0) = 1. If Y=a+bX, then the mgf of Y is given by If X,Y independent and Z = X+Y then, The last property can be generalized to a sum of n independent random variables. STA286 week 7

21 Uniqueness Theorem If a moment generating function mX(t) exists for t in an open interval containing 0, it uniquely determines the probability distribution. This can also be stated as follows: Let X and Y are two random variables with mgf MX(t) and MY(t), respectively. If MX(t) = MY(t) for all values of t, then X and Y have the same probability distribution. STA286 week 7

22 Example Find the mgf of X ~ N(μ,σ2) using the mgf of the standard normal random variable. Suppose, , independent. Find the distribution of X1+X2 using mgf approach. STA286 week 7

23 More on Normal Distribution
In general, If X1, X2,…, Xn i.i.d N(0,1) then X1+ X2+…+ Xn ~ N(0,n). If , ,…, then If X1, X2,…, Xn i.i.d N(μ, σ2) then Sn = X1+ X2+…+ Xn ~ N(nμ, nσ2) and STA286 week 7

24 Sum of Independent χ2(1) random variables
Recall: The Chi-Square density with 1 degree of freedom is the Gamma(½ , ½) density. If X1, X2 i.i.d with distribution χ2(1). Find the density of Y = X1+ X2. In general, if X1, X2,…, Xn ~ χ2(1) independent then X1+ X2+…+ Xn ~ χ2(n) = Gamma(n/2, ½). Recall: The Chi-Square density with parameter n is STA286 week 7

25 Characteristic Function
The characteristic function, cX, of a random variable X is defined by: The definition of the characteristics function is just like the definition of the mgf, except for the introduction of the imaginary number Using properties of complex numbers, we see that the characteristic function can also be written as The characteristics function unlike the mgf is always finite… STA286 week 7

26 Properties of Characteristic Function
The characteristics function has many nice properties similar to the mgf. In particular, it can be used to generate moments. Let X be any random variable with its first k moments finite. Then cX(0) = 1 etc. In general, Where is the kth derivative of cX with respect to s. Let X and Y be independent random variables. Then cX + Y (s) = cX(s) cY(s) STA286 week 7

27 Examples STA286 week 7


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