MATH 2421: Lecture 22

Date: 2024-11-20 11:27:40

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Conditional Expectation (cont.)

Conditional expectation
- we may also use this approach to compute probabilities
  - s.t. $X = I_{A}$ w/ $A$ being an event
    - then $E (I_{A}) = P (A)$
    - then $E (I_{A} ∣ Y = y) = P (A ∣ Y = y)$
  - we obtain: $P (A) = E (I_{A}) = E [E (I_{A} = y)]$ $= {\sum_{y} E (I_{A} ∣ Y = y) P (Y = 1) \int_{y} E (I_{A} ∣ Y = y) f_{y} (y) d y Y being discrete Y being continuous$
  - we obtain: $P (A) = E (I_{A}) = E [E (I_{A} = y)]$
    - ${\sum_{y} P (A ∣ Y = y) P (Y = 1) \int_{y} P (A ∣ Y = y) f_{y} (y) d y Y being discrete Y being continuous$
  - e.
- d
Example
- miner: trapped in a mine w/ 3 doors
  - first door: to a safety place after 3 hours
  - second door: back to same place after 5 hours
  - third door: back to same place after 7 hours
  - choosing each door: equally likely
    - and previous choice is forgotten
  - find: expected length of time, until the miner reach the safety
  - let $X$ : length of time until reaching the safety
    - $Y$ : door number he chooses $E (X ∣ Y = 1) E (X ∣ Y = 2) E (X ∣ Y = 3) P (Y = 1) E (X) = E_{Y} (E (X ∣ Y)) \frac{1}{3} E (X) E (X) = 3 = 5 + E (X) = 7 + E (X) = P (Y = 2) = P (Y = 3) = \frac{1}{3} = E (X ∣ Y = 1) P (Y = 1) + E (X ∣ Y = 2) P (Y = 2) + E (X ∣ Y = 3) P (Y = 3) = \frac{1}{3} \cdot 3 + \frac{1}{3} (5 + E (X)) + \frac{1}{3} (7 + E (X)) = 5 = 15$
- expectation of a random sum
  - suppose: $X_{1}, X_{2}, \dots$ are iid w/ common mean $μ$
  - let $N$ non-negative, integer valued r.v. independent of $X_{i}$
  - our interest: mean of $T = \sum_{k = 1}^{n} X_{k}$
    - where $T$ : taken to be zero when $N = 0$
  - solution $E (Y) = E (k = 1 \sum N X_{k}) = E_{N} [E (∣ N)] = n = 0 \sum \infty E (\sum X_{k} ∣ N = n) P (N = n) = n = 0 \sum \infty E (\sum n X_{k} ∣ N = n) P (N = n) = n = 0 \sum \infty E (\sum n X_{k}) P (N = n) = n = 0 \sum \infty k = 1 \sum n E (X_{k}) P (N = n) = n = 0 \sum \infty n μ P (N = n) = μ n = 0 \sum \infty n P (N = n) = μ E (N)$
- suppose: no. of people getting department store on a day: r.v. w/ $μ = 50$
  - amounts of money spent by individual customers: indep. r.v. w/ common mean $8$
  - suppose amount of money spent by a customer: independent of total no. customers entering
    - what is: expected amount of money spent on a given day?
- $X, Y$ : independent w/ pdf $f_{X}, f_{Y}$ . Compute $P (X < Y)$
  - conditioning on the value of $Y$ $P (X < Y) = \int_{- \infty}^{\infty} P (X < Y ∣ Y = y) f_{Y} (y) d y = \int_{- \infty}^{\infty} P (X < y ∣ Y = 1) f_{Y} (y) d y = \int_{- \infty}^{\infty} P (X < Y) f_{Y} (y) d y$
- let $U$ be u.r.v. and define conditional density is $X$ $p$
- $\frac{x ^{k} ( 1 - k ) ^{n - k}}{B ( k + 1 , n - k + 1 )}$

Conditional Variance

Conditional variance
- pre-req $var (X) = E [(X - E (X))^{2}] = E (X^{2}) - [E (X)]^{2}$ $var (X ∣ Y = y) var (X ∣ Y) = E [(X - E (X ∣ Y = y))^{2} ∣ Y = y] = E (X^{2} ∣ Y = y) - [E (X ∣ Y = y)^{2}] = h (y) = h (y) function of Y$
- conditional variance of $X$ given $Y = y$ $var (X ∣ Y) \equiv E [(X - E [X ∣ Y])^{2} ∣ Y]$
- notably: $var (X) = E [var (X ∣ Y)] + var (E [X ∣ Y])$
- proof
- 👨‍🏫 not often used, and its application is not expected
Example
- 👨‍🎓 (very complicated computation)

Moment Generating Function

Moment generating function
- moment generating function of r.v. $X$ , $M_{X}$ is defined as: $M_{X} (t) = E [e^{tX}] = {\sum_{X} e^{t x} p_{X} (x) \int e^{t x} f_{X} (x) d x if X discrete if X continuous$
- called as MFG as:
  - it generates moments of r.v. for $n \geq 0$
  - $E (X^{n}) = M_{X}^{(n)} (0)$
    - where $M_{X}^{(n)} (0) := \frac{d ^{n}}{d t ^{n}} M_{X} (t) ∣_{t = 0}$
- proof: by Taylor series expansion $E [e^{tX}] = E [k = 0 \sum \infty \frac{( tX ) ^{k}}{k !}] = k = 0 \sum \infty \frac{E ( tX ) ^{k}}{k !} = k = 0 \sum \infty \frac{E ( X ) ^{k}}{k !} t^{k}$
  - and $M_{X} (t) = \sum_{k = 0}^{\infty} \frac{M _{X}^{k} ( 0 )}{k !} t^{k}$
    - equating coefficient of $t^{n}$ , we get $M_{X}^{(n)} (0) = E (X^{n})$
  - 👨‍🏫 not all r.v. has a mgf
    - e.g. for cauchy distribution, $E (X^{n})$ does not exist for $n \geq 1$ , and $X$ has no mgf
      - 👨‍🏫 for advanced topics, people use complex numbers and find ways
- theorem: multiplicative property
  - if $X, Y$ independent:
  - $M_{X + Y} (t) = M_{X} (t) M_{Y} (t)$
- proof $E [e^{t (X + Y)}] = E [e^{tX} e^{t Y}] = E [e^{tX}] [e^{t Y}] = M_{X} (t) M_{Y} (t)$
- theorem: uniqueness property
  - let $X, Y$ be r.v. w/ their MGF $M_{X}, M_{Y}$
  - suppose there exists an $h > 0$ s.t.
    - $M_{X} (t) = M_{Y} (t), \forall t \in (- h, h)$
  - then $X, Y$ have the same distribution (i.e. $F_{X} = F_{Y}$ or $f_{X} = f_{Y}$ )

MATH 2421: Probability