Once we start using random variables as estimators, we will want to see how far the estimate is from a desired value. For example, we might want to see how far a random variable $X$ is from the number 10. That’s a function of $X$. Let’s call it $Y$. Then
which is not a linear function. To find $E(Y)$, we need a bit more technique. Throughout, we will assume that all the expectations that we are discussing are well defined.
This section is about finding the expectation of a function of a random variable whose distribution you know.
In what follows, let $X$ be a random variable whose distribution (and hence also expectation) are known.
Linear Function Rule
Let $Y = aX + b$ for some constants $a$ and $b$. In an earlier section we showed that
This includes the case where $a=0$ and thus $Y$ is just the constant $b$ and thus has expectation $b$.
Non-linear Function Rule
Now let $Y = g(X)$ where $g$ is any numerical function. Remember that $X$ is a function on $\Omega$. So the function that defines the random variable $Y$ is a composition:
This allows us to write $E(Y)$ in three equivalent ways:
On the range of $Y$
On the domain $\Omega$
On the range of $X$
As before, it is a straightforward matter of grouping to show that all the forms are equivalent.
The first form looks the simplest, but there’s a catch: you need to first find $P(Y=y)$. The second form involves an unnecessarily high level of detail.
The third form is the one to use. It uses the known distribution of $X$. It says that to find $E(Y)$ where $Y = g(X)$ for some function $g$:
- Take a generic value $x$ of $X$.
- Apply $g$ to $x$; this $g(x)$ is a generic value of $Y$.
- Weight $g(x)$ by $P(X=x)$, which is known.
- Do this for all $x$ and add. The sum is $E(Y)$.
The crucial thing to note about this method is that we didn’t have to first find the distribution of $Y$. That saves us a lot of work. Let’s see how our method works in some examples.
Example 1: $Y = \vert X-3 \vert$
Let $X$ have a distribution we worked with earlier:
x = np.arange(1, 6) probs = make_array(0.15, 0.25, 0.3, 0.2, 0.1) dist = Table().values(x).probability(probs) dist = dist.relabel('Value', 'x').relabel('Probability', 'P(X=x)') dist
Let $g$ be the function defined by $g(x) = \vert x-3 \vert$, and let $Y = g(X)$. In other words, $Y = \vert X - 3 \vert$.
To calculate $E(Y)$, we first have to create a column that transforms the values of $X$ into values of $Y$:
dist_with_Y = dist.with_column('g(x)', np.abs(dist.column('x')-3)).move_to_end('P(X=x)') dist_with_Y
To get $E(Y)$, find the appropriate weighed average: multiply the
P(X=x) columns, and add. The calculation shows that $E(Y) = 0.95$.
ev_Y = sum(dist_with_Y.column('g(x)') * dist_with_Y.column('P(X=x)')) ev_Y
Example 2: $Y = \min(X, 3)$
Let $X$ be as above, but now let $Y = \min(X, 3)$. We want $E(Y)$. What we know is the distribution of $X$:
To find $E(Y)$ we can just go row by row and replace the value of $x$ by the value of $\min(x, 3)$, and then find the weighted average:
ev_Y = 1*0.15 + 2*0.25 + 3*0.3 + 3*0.2 + 3*0.1 ev_Y
Example 3: $E(X^2)$ for a Poisson Variable $X$
Let $X$ have the Poisson $(\mu)$ distribution. You will see in the next chapter that it will be useful to know the value of $E(X^2)$. By our non-linear function rule,
This sum turns out to be hard to simplify. The term for $k=0$ is 0. In each term for $k \ge 1$, one of the $k$’s in the numerator cancels a $k$ in the denominator but the other factor of $k$ in the numerator remains. It would be so nice if that factor $k$ were $k-1$ instead, so it could cancel $k-1$ in the denominator.
This motivates the following calculation. No matter what $X$ is, if we know $E(X)$ and can find $E(X(X-1))$, then we can use additivity to find $E(X^2)$ as follows:
Let’s see if we can find $E(X(X-1))$ by applying the non-linear function rule.
We know that $E(X) = \mu$, so
Notice that $E(X^2) > (E(X))^2$. This is an instance of a general fact. Later in the course we will see why it matters.
For now, as an exercise, see if you can find $E(X(X-1)(X-2))$ and hence $E(X^3)$.