Expectation/Mean of RV Discrete Case: If X is a discrete random variable which can take the value x1, x2, x3…….. xn with respective probabilities p1, p2,………pn having then the expectation of X, denoted by E(x) is defined as E(x)=p1x1 + p2x2+……..+ pnxn = The expectation of discrete random variable X with PMF p(x) is defined as
Example in Discrete case
Expectation of RV Continuous Case: The expectation of continuous random variable X with PDF f (x) is defined as
Example in Continuous case
Expected value of function of RV Consider a random variable X with PDF or PMF f(x), If g(x) is a function of random variable X, then E[g(x)] is defined as
Example
Variance of a RV The variance is mean squared difference between each data point and the centre of the distribution measured by the mean.
Example of Variance
Conditional Expectation DISCRETE CASE: The conditional expectation or mean value of a function g(X, Y) given that Y= yi is defined by:
Conditional Expectation CONTINUOUS CASE: The conditional expectation or mean value of a function g(X, Y) given that Y= yi is defined by:
Example 1 Three coins are tossed. Let X denote the number of heads on the first two and Y denote the number of heads on the last two. Find: 1. The joint distribution of X and Y. 2. Marginal distribution of X and Y. 3. E(X) and E(Y) HHH HHT HTH THH HTT THT TTH TTT X 2 2 1 1 1 1 Y 2 1 1 2 1 1
Example 1 HHH HHT HTH THH HTT THT TTH TTT X 2 2 1 1 1 1 Y 2 1 1 2 1 1 X | Y 1 2 f(x) (Marginal) 1/8 1/8 1/4 JOINT DISTRIBUTION 1 1/8 2/8 1/8 1/2 2 1/8 1/8 1/4 f(y) 1/4 1/2 1/4 1
Example 1 X | Y 1 2 f(x) (Marginal) 1/8 1/8 1/4 1 1/8 2/8 1/8 1/2 2 1/8 1/8 1/4 f(y) 1/4 1/2 1/4 1
Example 2 X | Y 1 2 f(x) (Marginal) 1/8 1/8 1/4 1 1/8 2/8 1/8 1/2 2 1/8 1/8 1/4 f(y) 1/4 1/2 1/4 1 Find E[ Y | X=1]
Example 2 Find E[ Y | X=1]
Covariance Let X and Y be any two random variables and let E(X)=u1 and E(Y)=u2. Consider the mathematical expectation The number is known as covariance of X and Y and it is denoted by Cov (X, Y)
Chebyshev’s Inequality Let X is a random variable with mean ( μ ) and variance ( σ 2 ), then for any positive number k, we have or
Example Two unbiased dice are thrown. If X is the sum of the numbers showing up. Prove that P(|X-7|>=3)<=35/54. Compare this with the actual probability.
Example
Example Actual Probability=1/3=0.333 Chebyshev’s inequality=35/54=0.648 0.333<=0.648