Lesson 25 2D LOTUS
Theory
In this lesson, we calculate the expected value of a function of two random variables, \(E[g(X, Y)]\). For example, \(E[X + Y]\) and \(E[XY]\) are all examples of expected values that involve more than one random variable.
Again, the result is intuitive. Now that there are two random variables, the probabilities are given by the joint p.m.f. We use these probabilities to weight the possible values of \(g(X, Y)\).
Let’s apply Theorem 25.1 to the Xavier and Yolanda example from Lesson 18.
Example 25.1 (Xavier and Yolanda Revisited) In Lesson 18, we showed that the joint distribution of the number of times Xavier and Yolanda win is \[ \begin{array}{rr|cccc} & 5 & 0 & 0 & 0 & .0238 \\ & 4 & 0 & 0 & .0795 & .0530 \\ y & 3 & 0 & .0883 & .1766 & .0294 \\ & 2 & .0327 & .1963 & .0981 & 0 \\ & 1 & .0727 & .1090 & 0 & 0 \\ & 0 & .0404 & 0 & 0 & 0 \\ \hline & & 0 & 1 & 2 & 3\\ & & & & x \end{array}. \] Let’s calculate \(E[Y - X]\), the expected number of additional times that Yolanda wins, compared to Xavier, as well as \(E[XY]\), the expected product of the number of times they win.
Applying 2D LOTUS to the function \(g(x, y) = y-x\), we can calculate \(E[g(X, Y)] = E[Y - X]\). In the sum below, we will omit outcomes with probability 0. \[\begin{align*} E[Y - X] &= \sum_x \sum_y (y - x) f(x, y) \\ &\approx (0 - 0) .0404 + (1 - 0) .0727 + (1 - 1) .1090 \\ &\ \ \ + (2 - 0) .0327 + (2 - 1) .1963 + (2 - 2) .0981 \\ &\ \ \ + (3 - 1) .0883 + (3 - 2) .1766 + (3 - 3) .0294 \\ &\ \ \ + (4 - 2) .0795 + (4 - 3) .0530 + (5 - 3) .0238 \\ &= .947. \end{align*}\]
Of course, there is an easier to calculate this particular expected value. \(Y - X\) is just the number of wins in the last two bets, which follows a \(\text{Binomial}(n=2, p=18/38)\) distribution. From Appendix A.1, we know that the expected value of a binomial is \(np = 2 (18/38) \approx .947\), which matches the answer above.
We can calculate \(E[XY]\) in the same way, by applying 2D LOTUS to the function \(g(x, y) = xy\). \[\begin{align*} E[XY] &= \sum_x \sum_y xy f(x, y) \\ &= (0 \cdot 0) .0404 + (1 \cdot 0) .0727 + (1 \cdot 1) .1090 \\ &\ \ \ + (2 \cdot 0) .0327 + (2 \cdot 1) .1963 + (2 \cdot 2) .0981 \\ &\ \ \ + (3 \cdot 1) .0883 + (3 \cdot 2) .1766 + (3 \cdot 3) .0294 \\ &\ \ \ + (4 \cdot 2) .0795 + (4 \cdot 3) .0530 + (5 \cdot 3) .0238 \\ &= 4.11. \end{align*}\]Essential Practice
Two tickets are drawn from a box with \(N_1\) \(\fbox{1}\)s and \(N_0\) \(\fbox{0}\)s. Let \(X\) be the number of \(\fbox{1}\)s on the first draw and \(Y\) be the number of \(\fbox{1}\)s on the second draw. (Note that \(X\) and \(Y\) can only be 0 or 1.)
- Calculate \(E[XY]\) when the draws are made with replacement.
- Calculate \(E[XY]\) when the draws are made without replacement.
(Hint: You worked out the joint p.m.f. of \(X\) and \(Y\) in Lesson 18. Use it!)
You roll two fair, six-sided dice. Let \(X\) be the number on the first die. Let \(Y\) be the number on the second die. Calculate \(E[\max(X, Y)]\), the expected value of the larger of the two numbers. There are several ways you can do this. You should try to do this by applying 2D LOTUS to the joint distribution of \(X\) and \(Y\), which is extremely simple. To check your answer, you can use the p.m.f. of \(L = \max(X, Y)\) that you derived in Lesson 19.
Consider the following three scenarios:
- A fair coin is tossed 3 times. \(X\) is the number of heads and \(Y\) is the number of tails.
- A fair coin is tossed 4 times. \(X\) is the number of heads in the first 3 tosses, \(Y\) is the number of heads in the last 3 tosses.
- A fair coin is tossed 6 times. \(X\) is the number of heads in the first 3 tosses, \(Y\) is the number of heads in the last 3 tosses.
In all three scenarios, \(X\) and \(Y\) are both marginally \(\text{Binomial}(n=3, p=1/2)\). However, they have different joint distributions. In Lessons 18 and 19, you worked out the joint p.m.f.s.
Calculate \(E[X + Y]\) and \(E[XY]\) for each of the scenarios. Does \(E[X + Y]\) change, depending on the joint distribution? What about \(E[XY]\)?