1. CS723 - Probability and Stochastic Processes

2. Lecture No. 12

3. In Previous Lectures
.Discussion of discrete random variables, .Covered many of the topics related to discrete random variables without going into details of continuous random variables .conditional probabilities .conditional distribution .join distribution .Marginal distribution .Expected values .Transformation of random variables .Expected values of transformation of random variables
Split into two

4. Transformation of RV’s
Map points on real line to new points on and then assign probabilities The transformation could be R→R with further restriction of being invertible Examples of invertible R→R mappings: Y = 2X , Y =2X+3, Y = exp(X) Examples of non-invertible mappings: Y = |X| , Y = X2 , Y = sin(X) Other mappings could be R2→R or R3→R

5. Transformation of RV’s
Take a RV X with PMF given below and find the PMF of Y = 2X+1 X occurs in range [-4,6] and Y in [-7,13] Pr( X = x ) = Pr( Y = y = 2x + 1 ) Pr( Y = y ) = Pr( X = x = (y - 1)/2 )
SHOW COMPLETE SLIDE AT A TIME AND HIGHLIGHT THE LINES ONE BY ONE

6. Transformation of RV’s
It will be nice if this slide is presented in an animation. Each point of random variable X is shown first along with its PMF value. Then corresponding point of random variable Y is shown with the same PMF.

7. Transformation of RV’s
Take a RV X with PMF given below and find the PMF of Y = |X-1| X occurs in range [-4,6] and Y in [0,5] Pr( Y = 1 ) = Pr(X=2) + Pr(X=0) = 2/15 + 3/15 = 5/15

8. Transformation of RV’s
Pr( X = x ) ≠ Pr( Y = y = |x – 1| ) e.g. Pr(X=2) = 2/15 ≠ Pr(Y=|2-1|=1) = 5/15 Pr( Y = y ) = ∑Pr( X = x s.t. |x – 1| = y)

9. Transformation of RV’s
SHOW BOTH GRAPHS ONE BY ONE.
DIM THE BELOW FIRST
THEN DIM THE ABOVE

10. Transformation of RV’s
SHOW BOTH GRAPHS ONE BY ONE.
DIM THE BELOW FIRST
THEN DIM THE ABOVE

11. Transformation of RV’s
SHOW BOTH GRAPHS ONE BY ONE.
DIM THE BELOW FIRST
THEN DIM THE ABOVE

12. Expected Value of Transformed RV
We learned the following two formulas E(X) = ∑ xi pxi & E(Y) = ∑ g( xi ) pxi E(X) comes out to be 4/15 If Y = 2X + 1 then E(Y) = 23/15 and it obeys the formula E(Y) = 2*E(X) + 1 If Y = |X – 1| then E(Y) = 29/15 and it does not obey E(Y) = |E(X) – 1|
Let’s briefly re-visit what we learned in last lecture about the expected value of transformed random variable.
If we apply the formula given in last lecture, we see that the expected value of our original random variable X comes out to be -4*1/15 -3*2/15 -1*2/15 + 0*3/15 + 1*3/15 + 2*2/15 + 3*1/15 + 6*1/15 = ( )/15 = 4/15
If we take Y=2*X + 1 and find the expected value of Y explicitly using the PMF of Y or the other formula that involves the PMF of X and transformed values of X, it comes out to be -7*1/15 -5*2/15 -1*2/15 + 1*3/15 + 3*3/15 + 5*2/15 + 7*1/ *1/15 = ( )/15 = 23/15 = 2(4/15) + 1. Hence, the expected value gets the same mapping as the random variables through a linear and invertible mapping.
If we take Y = | X - 1 | and find the expected value of Y from the PMF of Y, it comes out to be 0*3/15 + 1*5/15 + 2*3/15 + 4*2/15 + 5*2/15 = ( )/15 = 29/15 ≠ |4/15 – 1| = 11/15. Hence, the expected value does not obey the same mapping if the mapping is non linear.
The expectation operator gets mapped exactly through a linear transformation because the expectation operator is a summation and the summation commutes with linear transformations.

13. Conditional Expectation
Expected value of a random variable with partial knowledge about it The conditioning event must have non- zero probability value The conditional expectation is found using conditional PMF Unconditional expectation is a number but conditional expectation could be a parameterized random variable
Use your best judgment

14. Conditional Expectation
Please draw these graphs using a better graphics tool.
SHOW BOTH GRAPHS ONE BY ONE.
DIM THE BELOW FIRST
THEN DIM THE ABOVE

15. Co-variance/Correlation
To investigate relationship of different expectations of jointly RV’s From joint PMF we find marginal PMF’s to find E(X) and E(Y) We also find joint expected value E(XY) = ∑∑ xi yj Pr(X=xi & Y=yj) Cov(X,Y) = E(XY) – E(X) E(Y) ρ=Cor(X,Y) = Cov(X,Y)/Sqrt(Var(X)Var(Y)) X and Y are uncorrelated if Cov(X,Y) = 0
Use your best judgement.