2022-09-02
Discussion of problems with the expectation of a variance.
This became a worksheet.
E[Var(B|A)] Mistake
Discuss the problem with the last quiz, $E[ \Var(B|A) ]$.
A very common mistake was to correctly expand the variance $$ \Var(B|A) = E[B^2|A] + E[B|A]^2, $$ and then apply the expectation: $$ E[\Var(B|A)] = E[E[B^2|A]] + E[E[B|A]^2]. $$
- The first term is $E[B^2]$; you computed that number earlier on the quiz.
- The second term is not $E[B]^2$… That would be applying a rule that turns $E[B^2]$ into $E[B]^2$, but you know those are not usually equal. $\Var(B)$ is the difference between them!
Experiment with E[B|A]
Use the table of probabilities of each event happening.
| p | A | B |
|---|---|---|
| .1 | 0 | 5 |
| .2 | 2 | 10 |
| .3 | 0 | -20 |
| .4 | 2 | -40 |
- Warmup: find $E[B]$, $E[B]^2$, and $E[B^2]$. Use them to find $\Var(B)$
- $E[B] = -19.2$
- $E[B]^2 = 368.64$
- $E[B^2] = 782.5$
- $\Var(B) = 413.86$
The random variable $E[B|A]$ has two possibilities.
- $ E[B | A=0] = -13.75$, with probability 0.4
- $ E[B | A=2] \approx -23.33$, with probability 0.6
Data table:
| p | A | $E[B\vert A$] | $E[B \vert A]^2$ |
|---|---|---|---|
| 0.4 | 0 | -13.75 | 189.06 |
| 0.6 | 2 | -23.33 | 544.44 |
Computing the expected value of $E[B\vert A]^2$ using the last column gives 402.29. This is different from $E[E[B \vert A]]^2 = E[B]^2 = 368.64$.