quartz/content/vault/econometrics/lectnot/L12 Continuous Joint Distributions.md

L12 Continuous Joint Distributions (WMS 5.1-8)

1. Joint Density
   • Compare discrete/continuous pdf and joint pdf
   • Warehouse stocks up to two pallets of cereal X and one pallet of cereal Y, with density f(x,y) = c(x + 2y);x\in\lbrack 0,2\rbrack,y\in\lbrack 0,1\rbrack.
   • Height of joint pdf represents likelihood of particular (x,y) pairs. Must integrate to one (double integral). 1 = \int_{x=0}^{2}\int_{y=0}^1c(x + 2y)dydx = \int_{x=0}^{2}(cx + c)dx = 4c requires c =\frac{1}{4}, or f(x,y) =\frac{1}{4}x +\frac{1}{2}y;x\in\lbrack 0,2\rbrack,y\in\lbrack 0,1\rbrack.

• Mode: since upward sloping in both dimensions, mode at (x,y) = (2,1)

2. Marginal densities

   • Analogous to margins of table in discrete joint distribution: total probability of particular realization of x is the sum of all joint probabilities of (x,y) pairs, with that particular x value, but any y value in domain.

   • f_{x}(x) =\int_{y=0}^{1}\frac{1}{4}(x + 2y)dy =\frac{1}{4}x +\frac{1}{4};x\in\lbrack 0,2\rbrack

   • f_{y}(y) =\int_{x=0}^{2}\frac{1}{4}(x + 2y)dx =\frac{1}{2} + y;y\in\lbrack 0,1\rbrack

   • Subscript simply distinguishes f_{x}(.5) from f_{y}(.5)

   • Moments: means, standard deviations

   1. $\mu_{x} = E(X) = \int_{x=0}^{2}xf_{x}(x)\text{dx}$ = \int_{x=0}^{2}x(\frac{1}{4}x +\frac{1}{4})dx =\frac{2}{3} +\frac{1}{2} =\frac{7}{6}
   2. $E(X^{2}) = \int_{x=0}^{2}x^{2}f_{x}(x)\text{dx}$ = \int_{x=0}^{2}x^{2}(\frac{1}{4}x +\frac{1}{4})dx = 1 +\frac{2}{3} =\frac{5}{3}
   3. V(X) = E(X^{2}) -\mu_{x}^{2} =\frac{5} {3} -(\frac{7}{6})^{2} =\frac{11}{36}
   4. \sigma_{x} = \sqrt{V(X)} =\sqrt{\frac{11}{36}}\approx .55
   5. $\mu_{y} = E(Y) = \int_{y=0}^{1}yf_{y}(y)\text{dy}$ = \int_{y=0}^{1}y(\frac{1}{2} + y)dy =\frac{1}{4} +\frac{1}{3} =\frac{7}{12}
   6. $E(Y^{2}) = \int_{y=0}^{1}y^{2}f_{y}(y)\text{dy}$ = \int_{y=0}^{1}y^{2}(\frac{1}{2} + y)dy =\frac{1}{6} +\frac{1}{4} =\frac{5}{12}
   7. V(Y) = E(Y^{2}) -\mu_{y}^{2} =\frac{5}{12} -(\frac{7}{12})^{2} =\frac{11}{144}
   8. \sigma_{y} = \sqrt{Y} =\sqrt{\frac{11}{144}}\approx 0.28
   9. Could also derive mode, median, cdf, percentiles of X or Y

• Independence requires f(x,y) = f_{x}(x)f_{y}(y) for all (x,y) pairs.
  1. X and Y not independent here, since f(x,y) =\frac{1}{4}(x + 2y)\neq(\frac{1}{4}x +\frac{1}{4})(\frac{1}{2} + y) (e.g. when (x,y) = (0,0))

3. Correlation

   • $E(\text{XY}) =\int_{x=0}^{2}\int_{y=0}^{1}\text{xyf}(x,y)\text{dydx}$ =\int_{x=0}^{2}\int_{y=0}^{1}\text{xy}\lbrack\frac{1}{4}(x + 2y)\rbrack dydx =\int_{x=0}^{2}(\frac{1}{8}x^{2} +\frac{1}{6}x)dx =\frac{1}{3} +\frac{1}{3} =\frac{2}{3}

   • $\sigma_{\text{xy}} = Cov(X,Y) = E(\text{XY}) -\mu_{x}\mu_{y}$ =\frac{2}{3} -(\frac{7}{6})(\frac{7}{12}) = -\frac{1}{72}

   • \rho =\frac{\sigma_{\text{xy}}}{\sigma_{x}\sigma_{y}} =\frac{-\frac{1}{72}}{(.55)(.28)}\approx - .09

4. Practice example: f(x,y) = c(1 - xy) for x,y\in\lbrack 0,1\rbrack

   • Find c: \int_{x=0}^{1}\int_{y=0}^{1}c(1 - xy)dydx =\frac{3}{4}c implies c =\frac{4}{3}

   • Find marginal densities f_{x}, f_{y}: f_{x}(x) =\int_{y=0}^1\frac{4}{3}(1 - xy)dy =\ldots =\frac{4}{3} -\frac{2}{3}x for x\in\lbrack 0,1\rbrack; symmetrically, f_{y}(y) =\frac{4}{3} -\frac{2}{3}y for y\in\lbrack 0,1\rbrack

   • Find means \mu_{x} and \mu_{y} and standard deviations \sigma_{x} and \sigma_{y}: \mu_{x} = E(X) = \int_{x=0}^1x(\frac{4}{3} -\frac{2}{3}x)dx =\ldots =\frac{4}{9} E(X^{2}) = \int_{x=0}^1x^{2}(\frac{4}{3} -\frac{2}{3}x)dx =\ldots =\frac{5}{18} \sigma_{x}^{2} =\frac{5}{18} -(\frac{4}{9})^{2} =\frac{13}{162} \sigma_{x} =\sqrt{\frac{13}{162}}\approx .283 Symmetrically, \mu_{y} =\frac{4}{9}, \sigma_{y}\approx .283

• Correlation \rho: E(\text{XY}) =\int_{x=0}^1\int_{y=0}^1\text{xy}\frac{4}{3}(1 - xy)dydx =\frac{5}{27} \sigma_{\text{xy}} = E(\text{XY}) -\mu_{x}\mu_{y}\approx\frac{5}{27} -(\frac{4}{9})^{2} = -\frac{1}{81}\approx - .012 \rho =\frac{\sigma_{\text{xy}}}{\sigma_{x}\sigma_{y}} =\frac{-\frac{1}{81}}{\sqrt{\frac{13}{162}}\sqrt{\frac{13}{162}}} = -\frac{2}{13}\approx - 0.154