Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 8 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii
- iii a
  - Solution iii

i

Solution

For an integrable discrete random variable, the conditional expectation given an event

B

of positive probability is

E [X ∣ B] = x \sum x P (X = x ∣ B) = x \sum x \frac{P ({ X = x } \cap B )}{P ( B )} .

The sum ranges over the possible values of

X

; integrability ensures absolute convergence.

ii

Solution

Using the indicator random variable, sum over the values of the discrete random variable

X

:

E [X 1_{B}] = \sum_{x} x P ({X = x} \cap B) = P (B) \sum_{x} x P (X = x ∣ B) .

Dividing by the positive probability

P (B)

proves $E [X ∣ B] = E [X 1_{B}] / P (B)$ . On

B^{c}

the product vanishes, so no additional contribution is omitted.

iii

Solution

The finite-partition form of the law of total expectation states

E [X] = j : P (B_{j}) > 0 \sum P (B_{j}) E [X ∣ B_{j}] .

Because the events form a partition of a sample space of the sample space,

\sum_{j} 1_{B_{j}} = 1

pointwise. Multiply by

X

and use linearity of expectation to obtain

E [X] = \sum_{j} E [X 1_{B_{j}}] = \sum_{j : P (B_{j}) > 0} P (B_{j}) E [X ∣ B_{j}] .

An event of zero probability contributes zero to the middle expression and is omitted from the final expression, so no undefined conditional expectation is multiplied by zero.

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version