IE 360: Design and Control of Industrial Systems I Lecture 10 Joint Random Variables Part 2 IE 360: Design and Control of Industrial Systems I References Montgomery and Runger Section 5-2 Copyright  2010 by Joel Greenstein

Expected Value of Joint RVs The expected value of joint rvs follows the same ideas as “regular” (aka “univariate”= “one variable”) rvs but you need to use the joint pmf/pdf Let X and Y be discrete random variables with joint pmf f(x, y). The expected value of some function of X and Y, denoted g(X, Y), is Let X and Y be continuous random variables with joint pdf f(x, y). The expected value of some function of X and Y, denoted g(X, Y), is g(X, Y) could be something like XY, or X/Y (as in “X divided by Y”) or other non-linear functions, or even just X or 3Y + 2 Rules of Expectation apply in this case still If two rvs are independent (their joint pmf/pdf equals the product of the marginals), then E[XY] = E[X]E[Y]

Discrete Joint RV Expectation Example I am notoriously clumsy. Let X be the number of times I drop the chalk and Y be the number of times I drop the eraser in a one-hour class. Observant students in the past have developed the following joint pmf to describe how often I drop the chalk and eraser when I teach. X=0 X=1 X=2 X=3 Y=0 0.1 0.3 0.26 Y=1 0.03 Y=2 0.02 Y=3 0.01 What is the expected number of times I drop the chalk? This is E[X], so we have g(X, Y) = X It is really the same as computing the expected value of X using the marginal distribution of X, since the sum of the joint pmf over the Y-values is the marginal pmf of X.

Continuous Joint RVs Example Consider joint continuous random variables X and Y, with pdf f(x, y) = e-x-y, x>0, y>0 Find E[X]; we have g(X, Y)=X More integration by parts!

Continuous Joint RVs Example 2 Consider joint continuous random variables X and Y, with pdf f(x, y) = e-x-y, x>0, y>0 Find E[Y]; we have g(X, Y)=Y This is just the same as E[X] but swapping X and Y, so we have E[Y] = 1 Find E[XY]; we have g(X, Y)=XY Yikes – a double dose of integration by parts!

Covariance Interpreting the covariance If two rvs are related through a joint pmf/pdf, we may ask: If one goes up, does the other go down? Do large changes in one correspond to large changes in the other? These questions are answered by the covariance, Cov(X, Y) = σXY Describes nature of linear relationship between two rvs Same idea for discrete or continuous joint rvs Definition: Cov(X, Y) = σXY = E[(X-μX)(Y-μY)] may be positive or negative!!! Easier computation: Cov(X, Y) = σXY = E[XY] – μXμY Interpreting the covariance If two rvs are independent, then Cov(X, Y) = 0 The reverse is not true. If the covariance between two random variables is zero, we cannot conclude that the two random variables are independent independence is a result of the pmf/pdf relationship described in the previous lecture An important factor is the sign of the covariance Positive indicates an increase in one rv is associated with an increase in the other Negative indicates a decrease in one rv is associated with an increase in the other

Covariance Example If E[X] = 1, E[Y] = -1, and E[XY] = 4, what is the covariance of X and Y? Cov(X, Y) = E[XY] – μXμY = E[XY] – E[X] E[Y] = 4 – (1)(-1) = 5 X and Y have positive covariance If E[W] = 10, E[Z] = -10, and E[WZ] = 400, what is the covariance of W and Z? Cov(W, Z) = E[WZ] – μWμZ = E[WZ] – E[W] E[Z] = 400 – (10)(-10) = 500 W and Z have positive covariance Are the relationships between X and Y and W and Z very different? Maybe , maybe not. The covariance is not unitless, so it is not meaningful to compare the covariances of different pairs of rvs

Correlation The correlation, a unitless extension of the covariance, can also be interpreted as a description of the linear relationship between two random variables Let X and Y be rvs with Cov(X, Y)= σXY and standard deviations σX and σY. The correlation, ρXY (pronounced “rho of X and Y”) is defined as follows: Now, we have a unitless number that ranges from -1 to +1: -1 ≤ ρXY ≤ 1 If the correlation is -1, we have perfectly negative correlation If the correlation is +1, we have perfectly positive correlation If the correlation is 0, we have no linear correlation x y ρXY = -1 x y ρXY = +1 x y ρXY = 0 x y ρXY = 0

Correlation Example Assume E[X] = ½ , E[Y] = 1/3, E[XY] = ¼, Var[X] = ¼ and Var[Y] = 1/12 What is the correlation of X and Y? First, compute the covariance Cov(X, Y) = E[XY] – μXμY = E[XY] – E[X] E[Y] = ¼ - ½(1/3) = 1/12 Now, compute the correlation X and Y are somewhat positively correlated

Related reading Montgomery and Runger, section 5-2 The captions of Figures 5-14, 5-15, and 5-16 should refer to Examples 5-21, 5-22, and 5-23, respectively. Some links that might help (maybe) Nice pictures of correlations are at http://en.wikipedia.org/wiki/Correlation http://cnx.org/content/m11248/latest/ Now you are ready to do HW 10