ev-continuous.Rmd

# Expected Value of Continuous Random Variables {#ev-continuous}
  
## Theory {-}

<iframe width="560" height="315" src="https://www.youtube.com/embed/hpGoz6-4CC8" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>

```{definition, name="Expected Value of a Continuous Random Variable"}
Let $X$ be a continuous random variable with p.d.f. $f(x)$. Then, the 
expected value of $X$ is defined as 
\begin{equation}
E[X] = \int_{-\infty}^\infty x \cdot f(x)\,dx.
(\#eq:ev-continuous)
\end{equation}
```
Compare this definition with the definition of expected value for a discrete random 
variable \@ref(eq:ev). We simply replaced the p.m.f. by the p.d.f. and 
the sum by an integral.

```{example ev-unif, name="Expected Value of the Uniform Distribution"}
Let $X$ be a $\text{Uniform}(a, b)$ random variable. What is $E[X]$?
  
First, the p.d.f. is 
\[ f(x) = \begin{cases} \frac{1}{b-a} & a \leq x \leq b \\ 0 & \text{otherwise} \end{cases}, \] 
which is non-zero only between $a$ and $b$. So, even though \@ref(eq:ev-continuous) says we should 
integrate from $-\infty$ to $\infty$, the integrand will only be non-zero between $a$ and $b$.

\begin{align*}
E[X] &= \int_a^b x \cdot \frac{1}{b-a} \,dx \\
&= \frac{1}{b-a} \left. \frac{x^2}{2}  \right]_a^b \\
&= \frac{1}{b-a} (b^2 - a^2) \frac{1}{2} \\
&= \frac{a + b}{2}.
\end{align*}

This is just the midpoint of the possible values of this uniform random variable. This is clearly the 
point where the p.d.f. would balance, if we put it on a scale.
```{r ev-continuous, echo=FALSE, engine='tikz', out.width='70%', fig.ext='png', fig.align='center', fig.cap='Expected Value of the Uniform Distribution', engine.opts = list(template = "tikz_template.tex")}
	\begin{tikzpicture}
	
		\draw [gray,<->] (-2, 0) coordinate -- (2, 0) coordinate;

    \fill[red!50] (-.3, -0.5) -- (0, 0) -- (.3, -0.5) -- cycle;
    \node[anchor=north,red] at (0, -0.5) {$\frac{a + b}{2}$};

    \node[anchor=north] at (-1, 0) {$a$};
    \node[anchor=north] at (1, 0) {$b$};
		
			\draw[thick] (-2, 0) coordinate -- (-1, 0) coordinate -- (-1, 1) coordinate -- (1, 1) coordinate -- (1, 0) coordinate -- (2, 0) coordinate;
	
	\end{tikzpicture}
```

```{example ev-expo, name="Expected Value and Median of the Exponential Distribution"}
Let $X$ be an $\text{Exponential}(\lambda)$ random variable. What is $E[X]$? Does the 
random variable have an equal chance of being above as below the expected value?
  
First, we calculate the expected value using \@ref(eq:ev-continuous) and the 
p.d.f. of the exponential distribution \@ref(eq:exponential-pdf). This is an 
exercise in integration by parts.
\begin{align*}
E[X] &= \int_0^\infty x \cdot \lambda e^{-\lambda x} \,dx \\
&=  -x e^{-\lambda x} \Big|_0^\infty - \int_0^\infty -e^{-\lambda x}\,dx \\
&= \ (-0 + 0) -  \underbrace{\frac{1}{\lambda} e^{-\lambda x} \Big|_0^\infty}_{0 - \frac{1}{\lambda}} \\
&= \frac{1}{\lambda}
\end{align*}

Now, let's calculate the probability that the random variable is below expected value.
\[ P(X < E[X]) = P(X < \frac{1}{\lambda}) = \int_0^{1/\lambda} \lambda e^{-\lambda x}\,dx = 1 - e^{-1} \approx .632. \]
The random variable does not have an 50/50 chance of being above or below its expected value.

The value that a random variable has an equal chance of being above or below is called its **median**. To 
calculate the median, we have to solve for $m$ such that 
\[ P(X < m) = 0.5. \]
(Equivalently, we could solve $P(X > m) = 0.5$. It also doesn't matter whether we use 
$<$ or $\leq$, since this is a continuous random variable, so $P(X = m) = 0$.)

Calculating the probability in terms of $m$, we have 
\[ 0.5 = P(X < m) = \int_0^m \lambda e^{-\lambda x}\,dx = 1 - e^{-\lambda m}. \]
Solving for $m$, we see that the median is $-\log(0.5) / \lambda$. (Note: $\log$ here is the natural logarithm, 
                                                                    base $e$.)
```{r, echo=FALSE, engine='tikz', out.width='70%', fig.ext='png', fig.align='center', fig.cap='Mean vs. Median of the Exponential Distribution', engine.opts = list(template = "tikz_template.tex")}
	\begin{tikzpicture}
  
    
    \draw[red!50,dashed,thick] (1, 0) -- (1, 2);
    \fill[red!50] (.8, -0.4) -- (1, 0) -- (1.2, -0.4) -- cycle;
    \node[anchor=east,red,rotate=90] at (1.01, -0.4) {$E[X]$};
    
    \fill [blue!40,smooth,samples=50,domain=0:.693] (0, 0) -- plot(\x,{2*exp(-(\x))}) -- (.693, 0);
    \draw[blue!40,dashed,thick] (.693, 0) -- (.693, 2);
    \node[anchor=east,blue,rotate=90] at (.65, 0) {median};
    \node[blue] at (0.33, 0.65) {{\small 50\%}};

		\draw [gray,->] (-0.1, 0) coordinate -- (3.5, 0) coordinate;
    \draw [gray,->] (0, -0.1) coordinate -- (0, 2.1) coordinate;
    \node[anchor=north] at (3.5, 0) {$x$};
    \node[anchor=east] at (0, 1.8) {$f(x)$};
    \draw [thick,smooth,samples=100,domain=0:3.5] plot(\x,{2*exp(-(\x))});
    
	  
	\end{tikzpicture}
```

Formulas for the variance of named continuous distributions can be found 
in Appendix \@ref(continuous-distributions).


## Essential Practice {-}

1. The distance (in hundreds of miles) driven by a trucker in one day is a continuous
random variable $X$ whose cumulative distribution function (c.d.f.) is given by: 
\[ F(x) = \begin{cases} 0 & x < 0 \\ x^3 / 216 & 0 \leq x \leq 6 \\ 1 & x > 6 \end{cases}. \]

    a. Calculate $E[X]$, the expected value of $X$.
    b. Calculate the median of $X$.
    c. Sketch a graph of the p.d.f., along with the locations of the expected value and median.

2. Suppose that an electronic device has a lifetime $T$ (in hours) that follows 
an $\text{Exponential}(\lambda=\frac{1}{1000})$ distribution.

    a. What is $E[T]$?
    b. Suppose that the cost of manufacturing one such item is $2. The manufacturer sells 
    the item for $5, but guarantees a total refund if the lifetime ends up being less than 900 hours. 
    What is the manufacturer's expected profit per item?