ELEC 303 – Random Signals

ELEC 303 – Random Signals Lecture 7 – Discrete Random Variables: Conditioning and Independence FarinazKoushanfar ECE Dept., Rice University Sept 15, 2009

Lecture outline Reading: Finish Chapter 2 Review Joint PMFs Conditioning Independence

Random Variables A random variable is a Real-valued function of an experiment outcome A function of a random variable defines another random variable We associate with each RV some averages of interest, such as mean and variance A random variable can be conditioned on an event or another random variable There is a notion of independence of a random variable from an event or from another

Discrete random variables • It is a real-valued function of the outcome of the experiments • can take a finite or infinitely finite number of values • A discrete random variable has an associated probability mass function (PMF) • It gives the probability of each numerical value that the random variable can take • A functionof a discrete random variable defines another discrete random variable (RV) • Its PMF can be found from the PMF of the original RV

Probability mass function (PMF) • Notations • Random variable: X • Experimental value: x • PX(x) = P({X=x}) • It mathematically defines a probability law • Probability axiom: xPX(x) = 1 • Example: Coin toss • Define X(H)=1, X(T)=0 (indicator RV)

Review: discrete random variable PMF, expectation, variance Probability mass function (PMF) PX(x) = P (X=x) x PX(x)=1

Expected value for functions of RV • Let X be a random variable with PMF pX, and let g(X) be a function of X. Then, the expected value of the random variable g(X) is given by • E[g(X)] = x g(x)pX(x) • Var(X) = E[(X-E[X])2] = x (x-E[X])2pX(x) • Similarly, the nth moment is given by • E[Xn]= xxnpX(x)

Properties of variance

Joint PMFs of multiple random variables • Joint PMF of two random variabels: pX,Y • PX,Y(x,y)=P(X=x,Y=y) • Calculate the PMFs of X and Y by the formula • PX(x)= yPX,Y(x,y) • PY(y)= XPX,Y(x,y) • We refer to PX and PY as the marginal PMFs

Tabular method • Assume Z=X+2Y • Find E[Z]? For computing marginal PMFs

Expectation

Variances

Example: Binomial mean and variance

More than two variables PX,Y,Z (x,y,z) = P(X=x,Y=y,Z=z) PX,Y (x,y) = z PX,Y,Z (x,y,z) PX(x) = y z PX,Y,Z (x,y,z) The expected value rule: E[g(X,Y,Z)] = x y z g(x,y,z)PX,Y,Z (x,y,z) http://www.coventry.ac.uk

Conditioning Conditional PMF of a RV on an event A PX|A(x)=P(X=x|A) = P({X=x} A)/P(A) P(A) = x P({X=x} A)  x PX|A(x) = 1

Example A student will take a certain test up to a max of n times, each time with a probability p of passing independent of the number of attempts Find the PMF of the number of attempts given that the student passes the test A={the event of passing} X is a geometric RV with parameter p and A={Xn} P(A) = {m=1 to n}(1-p)m-1p

Conditioning a RV on another PX|Y(x|y) = P(X=x|Y=y) PX|Y(x|y) = P(X=x,Y=y)/P(Y=y) = PX,Y(x,y)/PY(y) The conditional PMF is often used for the joint PMF, using a sequential approach PX,Y(x,y) = PY(y)PX|Y(x|y)

Mean and variance of Geometric Assume there is a probability p that your program works correctly (independent of how many times you write). Find the mean and variance of X, the number of tries till it works correctly? pX(k)=(1-p)k-1p, k=1,2,… E[X] = kk(1-p)k-1p Var(X) = k (k-E[X])2(1-p)k-1p

Mean and variance of Geometric • E[X|X=1]=1, E[X|X>1]=1+E(X) •  E[X] • E[X2|X=1]=1, E[X2|X>1]=E[(1+X)2]=1+2E[x]+E[X2] E[X2] = 1+2(1-p)E[X]/p

Independence Independence from an event P(X=x, A) = P(X=x)P(A) = PX(x) P(A), for all x P(X=x, A) = P(X=x and A) = PX|A(x)(A), PX|A(x)=PX(x), for all x Independence of random variables P(X=x,Y=y|A) =P(X=x|A)P(Y=y|A) for all x and y For two independent RVs: E[XY] = E[X]E[Y] Also, E[g(X)h(Y)] = E[g(X)]E[h(Y)]

Multiple RVs, sum of RVs Three RVs X, Y, and Z are said to be independent if PX,Y,Z (x,y,z) = PX(x)PY(y)PZ(z)

ELEC 303 – Random Signals