lotus2d.Rmd

# 2D LOTUS {#lotus2d}


## 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.

```{theorem lotus2d, name="2D LOTUS"}
Let $X$ and $Y$ be random variables with joint p.m.f. $f(x, y)$. Let $g$ be some function. Then,
\begin{equation}
E[g(X, Y)] = \sum_x \sum_y g(x, y) \cdot f(x, y).
(\#eq:lotus2d)
\end{equation}
```

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 \@ref(thm:lotus2d) to the Xavier and Yolanda example from 
Lesson \@ref(joint-discrete). 

```{example, name="Xavier and Yolanda Revisited"}
In Lesson \@ref(joint-discrete), 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 \@ref(discrete-distributions), 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 {-}

1.  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.)

    a. Calculate $E[XY]$ when the draws are made with replacement.
    b. 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 \@ref(joint-discrete). Use it!)

2. 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 \@ref(marginal-discrete).

3. 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 \@ref(joint-discrete) and \@ref(marginal-discrete), 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 $E[XY]$?