2022-09-02

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 \(E[E[B \vert A]]^2 = E[B]^2 = 368.64\) .