Chapter 10: Association between random variables

STAT 1010 - Fall 2022

Learning outcomes

By the end of this lesson you should:

• Understand the Sharpe ratio and know how to compute it to compare stocks that are not independent
• Understand and use a joint probability distribution function
• Know the 3 rules of independence
• Understand how covariance, correlation, and independence intertwine
• Explain how independent and identically distributed random variables impact upon expectation, variance, and standard deviation
• Compute expectation and variance of weighted sums

Revision 1

one of the main differences between this and chpt 6, is that here we are dealing with a population, before we were dealing with a sample.

Stress this throughout the class.

The Sharpe Ratio

write S(X) formula on the board

often mu and sigma are unknown parameters

Also, do we want the Sharpe ratio to be high or low?

\[S(X) = \frac{\mu_X - r_f}{\sigma_X}\]

• \(r_f\) is the return on a risk-free investment
• \(\mu\) is the mean of a random variable \(X\) that measures the performance of an investment
• \(\sigma\) is the standard deviation of a random variable \(X\) that measures the performance of an investment
• all inputs must be measured over the same time period (ie. yearly, monthly, …)

Compute the Sharpe ratio

have them go back to the previous slide and see what they got

Company	Random Variable	Mean per month	SD
Apple	A	2.45%	13.3%
McDonalds	M	1.14%	6.2%

assume \(r_f\) is the risk-free rate of interest is \(0.1\%\)

Compute the Sharpe ratio

work through it, then what if we don’t know mu and sigma could we find it if we had a pdf?

• \[\begin{aligned} S(A) &= \frac{\mu_A - r_f}{\sigma_A}\\ &= \frac{2.45 - 0.1}{13.3}\\ &= 0.177 \end{aligned}\]

• \[\begin{aligned} S(M) &= \frac{\mu_M - r_f}{\sigma_M}\\ &= \frac{1.14 - 0.1}{6.2}\\ &= 0.168 \end{aligned}\]

We prefer Apple because it has a higher Sharpe ratio \(0.177 > 0.168\)

Revision 2

find the mean, stdev, and use them to find the Sharpe ratio

put microsoft is y on the board and IBM is x

copy down the means and st deviation when they get them

Instead of knowing \(\mu\) and \(\sigma\) we have a pdf

	IBM stock	IBM stock	Microsoft	Microsoft
	\(x\)	\(P(X = x)\)	\(y\)	\(P(Y = y)\)
Increases	$5	0.11	$4	0.18
No change	0	0.80	0	0.67
Decreases	-$5	0.09	-$4	0.15

Sharpe ratio by hand

	IBM	IBM	IBM	IBM	IBM	IBM	IBM
	\(x\)	\(P(X = x)\)	\(\mu_X\)	\((x-\mu_X)^2 \cdot p_X(x)\)	\(Var(X)\)	\(sd(X)\)	Sharpe ratio
Increases	$5	0.11
No change	0	0.80
Decreases	-$5	0.09

Sharpe ratio by hand

	IBM	IBM	IBM	IBM	IBM	IBM	IBM
	\(x\)	\(P(X = x)\)	\(\mu_X\)	\((x-\mu_X)^2 \cdot p_X(x)\)	\(Var(X)\)	\(sd(X)\)	Sharpe ratio
Increases	$5	0.11	0.1	2.6411	4.99	2.23	0.03805123
No change	0	0.80	0.1	0.0080	4.99	2.23	0.03805123
Decreases	-$5	0.09	0.1	2.3409	4.99	2.23	0.03805123

Sharpe ratio by hand

	MCSFT	MCSFT	MCSFT	MCSFT	MCSFT	MCSFT	MCSFT
	\(y\)	\(P(Y =y)\)	\(\mu_Y\)	\((y-\mu_Y)^2 \cdot p_Y(y)\)	\(Var(Y)\)	\(sd(Y)\)	Sharpe ratio
Increases	$4	0.18
No change	0	0.67
Decreases	-$4	0.15

Sharpe ratio by hand

	MCSFT	MCSFT	MCSFT	MCSFT	MCSFT	MCSFT	MCSFT
	\(y\)	\(P(Y =y)\)	\(\mu_Y\)	\((y-\mu_Y)^2 \cdot p_Y(y)\)	\(Var(Y)\)	\(sd(Y)\)	Sharpe ratio
Increases	$4	0.18	0.12	2.709792	5.2656	2.29469	0.04575782
No change	0	0.67	0.12	0.009648	5.2656	2.29469	0.04575782
Decreases	-$4	0.15	0.12	2.546160	5.2656	2.29469	0.04575782

The Sharpe ratio - in R

Can an investor do better by diversifing the investment?

# Input the data and the pdf
stock <- tibble(x = c(5, 0, -5), 
                p_x = c(0.11, 0.8, 0.09), 
                y = c(4, 0, -4), 
                p_y = c(.18, .67, .15))


Sharpe_ratio <- # name this the Sharpe ratio
  stock %>% # use the data from above
  mutate(part_mean_x = x*p_x, # find mean for each part of x 
         part_mean_y = y*p_y, # now for y
         mean_x = sum(part_mean_x), # find mean x 
         mean_y = sum(part_mean_y),# now for y
         part_var_x = p_x * (mean_x - x)^2, # find var for each part of x 
         part_var_y = p_y * (mean_y - y)^2,# now for y
         var_x = sum(part_var_x), # find var of x 
         var_y = sum(part_var_y),# now for y
         sd_x = sqrt(var_x), # find sd of x 
         sd_y = sqrt(var_y)) %>% # now for y
  summarise(S_x = (mean_x - 0.015)/sd_x, # find Sharpe ratio of x with rf = 0.015
            S_y = (mean_y - 0.015)/sd_y) %>% # now for y
  slice(1)

Joint probability distributions

What is the expected value of X + Y?

When you see two random variables you first want to think independence.

Are these variables independent or not?

		\(X\)	\(X\)	\(X\)
		\(x = -5\)	\(x = 0\)	\(x = 5\)	\(p(y)\)
\(Y\)	\(y=4\)	\(0.00\)	\(0.11\)	\(0.07\)	\(0.18\)
\(Y\)	\(y=0\)	\(0.03\)	\(0.62\)	\(0.02\)	\(0.67\)
\(Y\)	\(y=-4\)	\(0.06\)	\(0.07\)	\(0.02\)	\(0.15\)
	\(p(x)\)	\(0.09\)	\(0.80\)	\(0.11\)	\(1\)

• What is the probability that \(x = 0\) and \(y =4\)? - \(11\%\)

• What is the probability that \(x = 5\) and \(y =4\)? - \(7\%\)

• What is the probability that \(x = 0\) and \(y =0\)? - \(62\%\)

• What outcome will never occur? - \(x = -5\) and \(y =4\)

Expected value of \(X + Y\)

If x=-5 then we subtract 5 from all y’s and multiply by \(p_{x, y}\) algebraically it all washes out

What about independence? We want to find the standard deviation to compute the Sharpe Ratio

		\(X\)	\(X\)	\(X\)
		\(x = -5\)	\(x = 0\)	\(x = 5\)	\(p(y)\)
\(Y\)	\(y=4\)	\(0.00\)	\(0.11\)	\(0.07\)	\(0.18\)
\(Y\)	\(y=0\)	\(0.03\)	\(0.62\)	\(0.02\)	\(0.67\)
\(Y\)	\(y=-4\)	\(0.06\)	\(0.07\)	\(0.02\)	\(0.15\)
	\(p(x)\)	\(0.09\)	\(0.80\)	\(0.11\)	\(1\)

• \(E(X+Y) = E(X) + E(Y)\)
• \(E(X+Y) = E(X) + E(Y) = 0.1 + 0.12 = 0.22\)

3 Rules of independence

So are X and Y independent?

If the probability of one event occuring has no impact on another event occuring, they are independent.

• If \(X\) and \(Y\) are independent then \(p(x,y) = p(x) \cdot p(y)\) for all pairs \((x,y)\)

• If \(p(x,y) = p(x) \cdot p(y)\) for all \((x, y)\) pairs, then \(X\) and \(Y\) are independent.

• It follows that if \(X\) and \(Y\) are independent, then \(E(XY) = E(X) \cdot E(Y)\)

Are \(X\) and \(Y\) independent?

find any p(x) X p(y) != p(x,y)

But we still need to find the stdev in order to find Sharpes ratio

		\(X\)	\(X\)	\(X\)
		\(x = -5\)	\(x = 0\)	\(x = 5\)	\(p(y)\)
\(Y\)	\(y=4\)	\(0.00\)	\(0.11\)	\(0.07\)	\(0.18\)
\(Y\)	\(y=0\)	\(0.03\)	\(0.62\)	\(0.02\)	\(0.67\)
\(Y\)	\(y=-4\)	\(0.06\)	\(0.07\)	\(0.02\)	\(0.15\)
	\(p(x)\)	\(0.09\)	\(0.80\)	\(0.11\)	\(1\)

• \(p_x(-5) \cdot p_y(4) = 0.18 \cdot 0.09 \neq 0.00\)

• \(p_x(-5) \cdot p_y(-4) = 0.09 \cdot 0.15 = 0.0135 \neq 0.06\)

• NO! NO! NO!

Covariance of Random Variables

last lesson (chpt 6) assumes that each \((x, y)\) pair appears once and are equally weighted

• \(cov(x, y) = \frac{(x_1 - \bar{x})(y_1 - \bar{y}) + (x_2 - \bar{x})(y_2 - \bar{y}) + \ldots + (x_n - \bar{x})(y_n - \bar{y})}{n-1}\)

• assumes each \((x, y)\) pair appears once

• \[Cov(X, Y) = E((X - \mu_X)(Y - \mu_Y))\]

Covariance of sum

because we square things and \((a + b)^2 = a^2 + b^2 + 2ab\) we have to add 2 of the covariance

• \((a+b)^2 = a^2 + 2ab +b^2\)
• it follows that
• \(Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y)\)

Correlation

what about our joint pdf? is it positively or negatively correlated? 3D plot

• \(corr(x, y) = \frac{cov(x, y)}{s_x \cdot s_y}\)

• \[\rho = Corr(X, Y) = \frac{Cov(X, Y)}{\sigma_X \cdot \sigma_Y}\]

As \(X\) increases, what happens to \(Y\)?

		\(X\)	\(X\)	\(X\)
		\(x = -5\)	\(x = 0\)	\(x = 5\)	\(p(y)\)
\(Y\)	\(y=4\)	\(0.00\)	\(0.11\)	\(0.07\)	\(0.18\)
\(Y\)	\(y=0\)	\(0.03\)	\(0.62\)	\(0.02\)	\(0.67\)
\(Y\)	\(y=-4\)	\(0.06\)	\(0.07\)	\(0.02\)	\(0.15\)
	\(p(x)\)	\(0.09\)	\(0.80\)	\(0.11\)	\(1\)

Independence & Cov

How does independence impact upon correlation and covariance?

\[ E((X-\mu_X)(Y-\mu_Y))\]

• \[\begin{aligned} Cov(X,Y) &= E((X-\mu_X)(Y-\mu_Y))\\ &= E(X - \mu_X)E(Y-\mu_Y)\\ &= 0 \end{aligned}\]

• If \(X\),\(Y\) are independent, then \(Cov(X, Y) = 0\)

• The opposite is not true (if \(Cov(X, Y) = 0\), then \(X\),\(Y\) are independent)

Independence & Var

\(Var(X) + Var(Y) = Var(X) + Var(Y) + 2Cov(X, Y)\)

• If \(X\), \(Y\) are independent, \(Cov(X, Y) = 0\), so

• \[Var(X+ Y) = Var(X) + Var(Y)\]

Sharpe ratio of a sum

\[ S(X + Y) = \frac{(\mu_X + \mu_Y) - 2r_f}{\sqrt{(Var(X+Y))}} \]

Sharpe ratio of sum - R

What if we want to invest in two subsequent days instead of investing in two stocks for 1 day?

# input x & y
x <- c(-5, 0, 5)
y <- c(4, 0, -4)

# input data
data <- bind_cols(expand_grid(x, y), 
                  probs = c(0, .03, .06, 
                            .11, .62, .07, 
                            .07, .02, .02), 
                  mu_x = rep(0.1, 9), 
                  mu_y = rep(0.12, 9), 
                  var_x = rep(4.99, 9), 
                  var_y = rep(5.2656, 9))

data %>% 
  mutate(cov_xy_part = (x - mu_x)*
           (y - mu_y)*probs, # multiply together
         cov_xy = sum(cov_xy_part), # sum them
         var_xy = var_x + var_y + 2*cov_xy) %>% # using the rule
  summarise(S_r = (mu_x + mu_y - 2*0.015)/
                      (sqrt(var_xy))) %>% # find the Sharpe value
  slice(1) # only the first

# A tibble: 1 × 1
     S_r
   <dbl>
1 0.0497

• The Sharpe ratio for \(X\) is \(0.038\), for \(Y\) it’s \(0.046\), investing in both gives a better return \(0.050\)

Double the investment

• Invest twice as much for one day
• Invest in one stock for two subsequent days
• Which is better? Why?

Double for one day

\[S(2X) = \frac{2\mu_X - 2r_f}{\sqrt{Var(2X)}}\]

• \(\mu_X = 0.1\)

• \(Var(X) = 4.99\)

• \(r_f = .015\)

• \[\begin{aligned} &= \frac{2 \cdot 0.1 - 2 \cdot 0.015}{\sqrt{4 \cdot 4.99}}\\ &= \frac{.2 - 0.03}{\sqrt{19.96}}\\ &= 0.038 \end{aligned}\]

• This is the same as before, how do we compute if we invest for two subsequent days?
  

IID

If we invested in a stock on 2 subsequent days instead of investing in 2 stocks on one day the return on those two days are:

independent and identically distributed (IID)

• identically distributed - the outcomes on each day are likely to be different, but the probability of the outcomes is the same

• independent - very common assumption for stocks (part of the reason for the 2008 financial crises)

Addition rules for IID variables

Important

If \(n\) random variables \((X_1, X_2, ..., X_n)\) are iid with mean \(\mu_X\) and standard deviation \(\sigma_X\), then

\[ E(X_1 + X_2 + ... + X_n) = n \cdot \mu_X \] \[ Var(X_1 + X_2 + ... + X_n) = n \cdot \sigma^2_x \] \[ SD(X_1 + X_2 + ... + X_n) = \sqrt{n} \sigma_X \]

Sharpe ratio - two days

if you decide to leave money in the bank, you may end up with a constant

\[ S(X_1 + X_2) = \frac{2\mu_X - 2r_f}{\sqrt{2 \sigma^2_x}}\]

• \[\begin{aligned} &= \frac{0.20 - 0.03}{\sqrt{9.98}}\\ &= 0.054 \end{aligned}\]
• you expect more variance if you have two stock for one day b/c anything that happens that day is magnified
• if you have one stock for two days you reduce the variance

Weighted sums

If we decide to leave money in the bank, there will be interest that accrues. How can we include this in our model?

Important

\[E(aX + bY + c) = aE(X) + bE(Y) + c\]

\[Var(aX + bY + c) = a^2Var(X) + b^2Var(Y) + 2abCov(X, Y)\]

Expectation of weighted sum

\[ E(2X + 4Y + 0.06)\]

• • \[\begin{aligned} &= 2E(X) + 4E(Y) + 0.06\\ &= 2(0.10) + 4(0.12) + 0.06\\ &= \$0.74 \end{aligned}\]

Variance of weighted sum

\[ Var(2X + 4Y + 0.06)\]

• \[\begin{aligned} &= 2^2Var(X) + 4^2Var(Y) + 2 \cdot(2 \cdot 4) \cdot Cov(X,Y)\\ &= 4(4.99) + 16(5.27) + 16(2.19)\\ &= 139.32 \end{aligned}\]

Your turn

Click here or the qr code below