MATH 2421: Lecture 11

Date: 2024-10-09 12:02:10

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❓Questions

Notes

Hypergeometric Random Variable (cont.)

Practice
- purchaser of electrical component: buys in lots of size 10
  - policy: inspect 3 from a lot
  - and all must be non-defective
  - w/ 30% of lots having 4 defective components
  - and 70% w/ 1 defective component
  - what proportion of lots: does purchaser reject?
  - $X \sim H (n, N, m)$
    - $N = 10$
    - $n = 3$
  - $m = 1∣∣4$
    - $P (m = 1) = 0.7$
    - $P (m = 4) = 0.3$ $P (acceptance) P (rejection) = P (X = 0) = P (X = 0∣ m = 4) P (m = 4) + P (X = 0∣ m = 1) P (m = 1) = \frac{( 0 4 ) ( n - 0 10 - 4 )}{( 3 10 )} \cdot 0.3 + \frac{( 0 1 ) ( n - 0 10 - 1 )}{( 3 10 )} \cdot 0.7 = 0.54 = 1 - 0.54 = 0.46$

Expected Value of Sum of Random Variables

Expected value of sum of random variables
- 👨‍🏫 more discussion in Ch. 7; important!
- for a r.v. $X$ , let $X (s)$ : val. of $X$ when $s \in S$ being output
- if $X, Y$ : both r.v. $⟹$ their sum is also r.v.
- $Z = X + Y$ : also r.v.
  - $Z (s) = X (s) + Y (s)$
  - r.v.: map from sample space to real line
- theorem: let $p (x) = P ({s})$ : probability that $s$ : output of experiment
  - then $E [X] = s \in S \sum X (s) p (s)$
- proof
  - suppose, for distinct values of $X$ are $x_{i} \geq 1$
  - for each $i$ : $S_{i}$ : event $X = x_{i}$
    - i.e. $S_{i} = {s : X (s) = x_{i}}$ $E [X] = i \sum x_{i} P {X = x_{i}} = i \sum x_{i} P (S_{i}) = i \sum x_{i} s \in S_{i} \sum p (s) = i \sum s \in S_{i} \sum x_{i} p (s) = i \sum s \in S_{i} \sum X (s) p (s) = s \in S \sum X (s) p (s)$
- ⭐ linearity of expectation
  - for r.v. $X_{1}, X_{2}, \dots, X_{n}$ (no independence req.) $E [i = 1 \sum n X_{i}] = i = 1 \sum n E [X_{i}]$
- proof $Z E (Z) := i = 1 \sum n X_{i} = s \in S \sum Z (s) p (s) = s \in S \sum (i = 1 \sum n X_{i}) p (s) = s \in S \sum (i = 1 \sum n X_{i} p (s)) = i = 1 \sum n (s \in S \sum X_{i} p (s)) = i = 1 \sum n E [X_{i}]$
Practice
- suppose: experiment consists of flipping a coin 5 times
  - result: only heads & tails
  - $X$ : no. of heads in first 3 flips
  - $Y$ : no. of heads in last 2 flips
  - then $Z = X + Y$ - for any combination
- suppose: two independent flip of a coin
  - head w/ probability of $p$ maid
  - $X$ : no. of heads obtained
  - $P (X = 0) = P (t, t) = (1 - p)^{2}$
  - $P (X = 1) = P (h, t) + P (t, h) = 2 p (1 - p)$
  - $P (X = 2) = P (h, h) = p^{2}$
  - thus $E [X] = 1 \cdot 2 p (1 - p) + 2 \cdot p^{2} = 2 p$
  - as well as $E [X] = X (h, h) p^{2} + X (h, t) p (1 - p) + X (t, h) (1 - p) p + X (t, t) (1 - p)^{2} = 2 p^{2} + p (1 - p) + (1 - p) p = 2 p$
- find: expected total no. of success from $n$ trials
  - when trial $i$ : success w/ probability $p_{i}$
  - let $x_{i} = {10 i -th: success otherwise$ $E (X) = E (i = 1 \sum n x_{i}) = i = 1 \sum n E (x_{i}) = i = 1 \sum n p_{i}$
  - special case: $\forall p_{i} = p$ and trials independent:
    - then $X \sim Bin(n,p)$
    - $E (X) = \sum_{i = 1}^{n} p = n p$
- $n$ identical & independent trials
  - each trial: yields success w/ same $p$
    - $X_{i} = 1$ if success
  - proof
    - show: $\sum_{i = 1}^{n} X_{i} \sim Bin (n, p)$
    - by def. $X = \sum_{i = 1}^{n} X_{i}$ : total no. of success in $n$ independent $Bernoulli (p)$
    - thus $\sum_{i = 1}^{n} X_{i} \sim Bin (n, p)$
    - ⭐⭐ w/ same $p$ , then always
      - $X = \sum_{i = 1}^{n} X_{i}$
        
        where $X \sim Be (p)$
        
        and $x_{i}$ all independent
- for $X \sim Bin (n, p), var (X) = ?$ $E (X) var (X) E (X^{2}) E (X_{i}^{2}) i \neq = j, x_{i} x_{j} E (x_{i} x_{j}) ⟹ E (X^{2}) var (X) = n p = E (X^{2}) - (E (X))^{2} = E [(i = 1 \sum n x_{i})^{2}] = E [((i = 1 \sum n x_{i}) \cdot (j = 1 \sum n x_{j}))^{2}] = E (i = 1 \sum n x_{i}^{2} + i = 1 \sum n j \neq = 1 \sum n x_{i} \cdot x_{j}) = E [i = 1 \sum n x_{i}^{2}] + E i = 1 \sum n j \neq = 1 \sum n x_{i} \cdot x_{j} = i = 1 \sum n E (x_{i}^{2}) + i = 1 \sum n j \neq = 1 \sum n E (x_{i} \cdot x_{j}) = E (X_{i}) = p = {10 if x_{i} = 1, x_{j} = 1 = 1 \times P (X_{i} = 1, X_{j} = 1) + 0 \times P (otherwise) = P (X_{i} = 1) P (X_{j} = 1) = p^{2} = i = 1 \sum n p + i = 1 \sum n j \neq = 1 \sum n p^{2} = n p + n (n - 1) p^{2} = E (X^{2}) - (E (X))^{2} = n p + n (n - 1) p^{2} - (n p)^{2} = 0 // as X_{i} \in {0, 1}$

Continuous Random Variables: Introduction

Continuous r.v.
- r.v. w/ range being an interval over real line
  - weight / time / length ...
  - considered separate as:
    - 👨‍🏫 you can't get exactly 30 cm long ruler, etc.
- for a continuous r.v. $X$
  - $\forall t \in R, P (X = t) = 0$
  - and thus $P (a < X < b) = P (a \leq X \leq b)$
    - and $P (a < X < b) = P (a < X < b) + P (X = a) + P (X = b) = P (a \leq X \leq b)$
- for cont. r.v. $X$ , w/ property s.t.
  - $f_{X}$ : non-negative function, $\forall x \in R$ $P (X \in B) = \int_{B} f_{X} (x) d x$
  - then: $f_{X}$ : probability density function (pdf)
- for all $B = [a, b]$ $P (a \leq X \leq b) = \int_{a}^{b} f_{X} (x) d x = F_{X} (b) - F_{X} (a)$
- $P (X = a)$ as: $P (X = a) = P (a \leq X \leq a) = \int_{a}^{a} f (x) d x = 0$
- distribution function (cdf): defined by $F_{X} (x) = P (X \leq x) = \int_{- \infty}^{x} f_{X} (t) for x \in R$
  - 👨‍🏫 same as discrete!
- w/ fundamental theorem of calculus:
  - $F_{X}^{'} (x) = f_{X} (x), x \in R$
  - density: derivative of cumulative distribution function $P (x - \frac{ε}{2} \leq X \leq x + \frac{ε}{2}) = \int_{x - ε /2}^{x + ε /2} f_{X} (x) d x \approx ε f (x)$
  - (think of it as a area of small slice of rectangle)
  - $f (x)$ : measure of how likely that r.v. will be near $x$
    - $\neq =$ probabilities
    - $f (x) \in [0, \infty)$

MATH 2421: Probability