quartz/content/vault/econometrics/lectnot/L13 Conditional Distributions.md

L13 Conditional Distributions (WMS 5.3, 11)

1. Recall distribution of cell phone use and car accidents:

	Y = 0	Y = 1	Y = 2
X = 0	.60	.08	.02
X = 1	.10	.12	.08

• Recall \mu_{x} = .3,\sigma_{x}\approx .458,\mu_{y} = .4,\sigma_{y}\approx .663,\rho\approx .527

2. Conditional probability

   • Cell phone use among those with two accidents P(X=1|Y = 2) =\frac{.08}{.10} = .80 versus those with no accidents P(X=1|Y = 0) =\frac{.10}{.70}\approx 0.14

   • Practice: find P_{y}(0|X = 0) =\frac{.60}{.7}\approx .86, P_{y}(1|X = 0) =\frac{.08}{.7} = .11,$P_{y}(2|X = 0) =\frac{.02}{.7}\approx .03$, P_{y}(0|X = 1) =\frac{.10}{.3}\approx .33, P_{y}(1|X = 1) =\frac{.12}{.3} = .40, P_{y}(2|X = 1) =\frac{.08}{.3}\approx .27

3. Conditional distribution

   • P(y|X = 0) and P(y|X = 1) are legitimate distribution functions (numerators sum to denominator)

   • Plot, and compare with P(y)

   • Conditional mean

   1. E(Y|X = 1) =\sum\text{yP}(X = 1) \approx 0(.33) + 1(.40) + 2(.27) = .94
   2. Practice E(X = 0)\approx 0(.86) + 1(.11) + 2(.03) = .17
   • Average number of car accidents is higher for cell phone users than for non-users. This still doesn't imply causation!
   • Conditional standard deviation
     1. Just like V(Y) = E(Y^{2}) -\lbrack E(Y)\rbrack^{2}, $V(Y|X = x) = E(Y^2|X = x) -\lbrack E(Y|X = x)\rbrack^{2}$ 1. (If time) Example : E(Y^2|X = 1) \approx 0^{2}(.33) + 1^{2}(.40) + 2^{2}(.27) = 1.21, so In this case, V(Y|X = 1)\approx 1.21 - {.94}^{2}\approx \sqrt{0.326}, and \sigma_{y|X = 1}\approx\approx 0.57 By a similar derivation, \sigma_{y|X = 0}\approx 0.41; cell phone use increases variance.

4. In an effort to establish causation, could find P(x,y|Z = z) =\frac{P(x,y,z)}{P_{z}(z)} or f(Z = z) =\frac{f(x,y,z)}{f_{z}(z)}, and then \rho_{xy|z} = Corr(X,Y|Z = z). For example, find correlation between cell phone use and car accidents among teenagers.

5. Continuous densities

   • Recall joint density of cereal inventory, f(x,y) =\frac{1}{4}x +\frac{1}{2}y;x\in\lbrack 0,2\rbrack,y\in\lbrack 0,1\rbrack

   • Recall marginal densities f_{x}(x) =\frac{1}{4}x +\frac{1}{4};x\in\lbrack 0,2\rbrack and f_{y}(y) =\frac{1}{2} + y;y\in\lbrack 0,1\rbrack, means \mu_{x} =\frac{7}{6}, \mu_{y} =\frac{7}{12}, standard deviations \sigma_{x}\approx .55, \sigma_{y}\approx 0.28

   • Conditional density f_{x}(x|Y = y) =\frac{f(x,y)}{f_{y}(y)} =\frac{\frac{1}{4}x +\frac{1}{2}y}{\frac{1}{2} + y} =\frac{x + 2y}{2 + 4y};x\in\lbrack 0,2\rbrack. For example, f_{x}(x|Y = 0) =\frac{x}{2};x\in\lbrack 0,2\rbrack

   • Conditional mean and standard deviation E(X|Y = 0) = x(\frac{x}{2})dx =\lbrack\frac{1}{6}x^{3}\rbrack_{0}^{2} =\frac{8}{6} =\frac{4}{3} E(X^2|Y = 0) = x^{2}f_{x}(0)dx = x^{2}(\frac{x}{2})dx =\lbrack\frac{1}{8}x^{4}\rbrack_{0}^{2} = 2 V(X|Y = 0) = E(Y = 0) -\lbrack E(Y = 0)\rbrack^{2} = 2 -(\frac{4}{3})^{2} =\frac{2}{9}. \sigma_{x|Y = 0} = Thus, when Y = 0: density of X is steeper, mean of X is higher, variance is lower.

• More generically, for arbitrary y value, E(X|Y = y) = \int xf_{x}(x|y)dx = \int_0^2x(\frac{x + 2y}{2 + 4y})\text{dx} $=\frac{1}{2 + 4y}\int_0^2(x^{2} + 2xy)dx =\frac{1}{2 + 4y}\lbrack\frac{1}{3}x^{3} + x^{2}y\rbrack_{0}^{2} =\frac{1}{2 + 4y}(\frac{8}{3} + 4y) =\frac{4 + 6y}{3 + 6y}$ For example, E(X|y = 0) =\frac{4}{3} as before Practice: E(X|y = 1) =\frac{10}{9} E(X^2|Y = y) = \int x^{2}f_{x}(x|y)\text{dx} = \int_0^2x^{2}(\frac{x + 2y}{2 + 4y})dx =\ldots =\frac{6 + 8y}{3 + 6y} V(X|Y = y) = E(X^2|Y = y) -\lbrack E(X|Y = y)\rbrack^{2} =(\frac{6 + 8y}{3 + 6y}) -(\frac{4 + 6y}{3 + 6y})^{2} \sigma_{x|Y = y} =\sqrt{(\frac{6 + 8y}{3 + 6y}) -(\frac{4 + 6y}{3 + 6y})^{2}}. For example, \sigma_{x|Y = 0} = \sqrt{\frac{6}{3}-(\frac{4}{3})^2}=\sqrt{\frac{2}{9}} as before, \sigma_{x|Y = 1} =\sqrt{(\frac{14}{9})-(\frac{10}{9})^2}=\sqrt{\frac{26}{81}}\approx\frac{5}{9} Thus, when Y = 1, density of X is less steep, mean is lower, variance is higher.

6. With three variables, can derive joint distribution of X and Y conditional on Z

   • Israel covid data
     1. When covid Delta variant hit, Israeli hospitals filled up with covid patients. 60% of these patients had already been vaccinated.
     2. Natural (but erroneous) conclusion: vaccines make things worse, not better!
     3. Nearly 80% of Israelis over age 12 were vaccinated against covid, so if "random draw," 80% of hospital patients should have been vaccinated; lower rate means treatment helped.