19 Continuous: joint & conditional models

19.1 Discussion

Let \(X\) and \(Y\) denote the proportions of time in one work week that employees A and B spend on task, respectively. Suppose the joint behavior of \(X\) and \(Y\) can be modeled by \[f_{X,Y}(x,y) = 2x \;\; \text{ for } 0 \le x \le 1, \; 0 \le y \le 1 \; .\]

This pdf is a theoretical representation of the reality we observe day after day:





EXAMPLE 1: joint intuition

  1. How can we tell if \(f_{X,Y}(x,y)\) is a valid joint pdf?
  2. How can we use the joint pdf to calculate the probability that, in a given week, both A and B spend less than 50% of their time on task?



EXAMPLE 2: marginal intuition

Use the plot of the joint pdf to develop an understanding about the marginal behaviors of \(X\) and \(Y\). How might we specify these marginal models?





Joint models - Continuous RVs

The simultaneous behavior of \(X\) and \(Y\) is described by their joint pdf \[f_{X,Y}(x,y)\] where

  • \(f_{X,Y}(x,y) \ge 0\) for all \(x,y\)
  • \(\int \int f_{X,Y}(x,y) dx dy = 1\)
  • \(f_{X,Y}(x,y)\) measures the density, NOT probability, at \((x,y)\)
  • Given the joint behavior of \(X\) & \(Y\), we can find their marginal behaviors using the Law of Total Probability!: \[\begin{split} f_X(x) & = \int f_{X,Y}(x,y)dy \\ f_Y(y) & = \int f_{X,Y}(x,y)dx \\ \end{split}\]





EXAMPLE 3: conditional

What’s your intuition about the conditional behavior of \(X\) when \(Y=0.5\) (left) is known? What about the conditional behavior of \(Y\) when \(X=0.5\) (right) is known? How might we specify the conditional pdfs of \(X\) and \(Y\)?





Conditional models - Continuous RVs

The conditional behavior of \(Y\) given that we observe \(X=x\) is described by the conditional pdf \[f_{Y|X=x}(y|x)\] where

  • \(f_{Y|X=x}(y|x) \ge 0\) for all \(x,y\)
  • \(\int f_{Y|X=x}(y|x) dy = 1\)
  • \(f_{Y|X=x}(y|x)\) measures the conditional density, NOT probability, at \(y\)
  • Given the joint & marginal behaviors of \(X\) & \(Y\), we can find their conditional behaviors: \[\begin{split} f_{Y|X=x}(y|x) & = \frac{f_{X,Y}(x,y)}{f_X(x)} \end{split}\]

\(X\) and \(Y\) are independent if and only if \[\begin{split} f_{Y|X=x}(y|x) & = f_Y(y) \\ f_{X,Y}(x,y) & = f_X(x)f_Y(y) \\ \end{split}\]

NOTE: These statements are equivalent. You only need to check one of them!





19.2 Exercises

  1. Joint
    1. Prove that \(f_{X,Y}(x,y) = 2x\) for \(0 \le x \le 1, \; 0 \le y \le 1\) is a valid joint pdf.
    2. Use the joint pdf to calculate the probability that, in a given week, both A and B spend less than 50% of their time on task.


    Solution

    1. \(f_{X,Y}(x,y) \ge 0\)
      \(\int_0^1\int_0^1 2x dxdy = \int_0^1 1 dy = 1\)

    2. \(\int_0^{0.5}\int_0^{0.5} 2x dxdy = \int_0^{0.5} 0.25 dy = 0.125\)



  1. Marginal
    Derive the marginal pdfs of \(X\) and \(Y\), \(f_X(x)\) and \(f_Y(y)\), from the joint pdf \(f_{X,Y}(x,y) = 2x\) for \(0 \le x \le 1, \; 0 \le y \le 1\). These are plotted below:


    Solution

    \[\begin{split} f_X(x) & = \int_0^1 2x dy = 2x \int_0^1 1 dy = 2x \\ f_Y(y) & = \int_0^1 2x dx = 1 \\ \end{split}\]



  1. Conditional
    Some conditional intuition is provided in the plots below.

    1. Derive and sketch the conditional pdf of \(X\) given \(Y=0.5\), \(f_{X|Y=0.5}(x|y=0.5)\).

    2. For any given \(y\), derive and sketch the conditional pdf of \(X\) given \(Y=y\), \(f_{X|Y=y}(x|y)\).

    3. Are \(X\) and \(Y\) independent RVs? Explain.


Solution

  1. .
    \[\begin{split} f_{X|Y=0.5}(x|y=0.5) & = \frac{f_{X,Y}(x,0.5)}{f_Y(0.5)}\\ & = \frac{2x}{1}\\ & = 2x \\ \end{split}\]

  2. .
    \[\begin{split} f_{X|Y=y}(x|y) & = \frac{f_{X,Y}(x,y)}{f_Y(y)}\\ & = \frac{2x}{1}\\ & = 2x \\ \end{split}\]

  3. Yes. \(f_{X|Y}(x|y) = f_X(x)\). Info about \(Y\) doesn’t change our understanding of \(X\).



  1. Independence
    Are \(X\) and \(Y\) independent?
    1. \(f_{X,Y}(x,y) = 6e^{-2x-3y}\) for \(x,y>0\)
    2. \(f_{X,Y}(x,y) = 1\) for \(x \in (0,1)\) and \(y \in (0,1)\)
    3. \(f_{X,Y}(x,y) = 1\) for \(0 < y < x < \sqrt{2}\)


    Solution

    1. Yes. \(f_{X,Y}(x,y) = 6e^{-2x-3y} = 2e^{-2x} * 3e^{-3y} = f_X(x)f_Y(y)\)
    2. Yes. \(f_{X,Y}(x,y) = 1 = 1*1 = f_X(x)f_Y(y)\)
    3. No. The fact that the possible values of \(X\) are constrained by the values of \(Y\) tells us that information about \(Y\) will certainly change our understanding of \(X\).