MATH 2421: Lecture 12

Date: 2024-10-14 11:50:44

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Continuous Random Variables: Introduction (cont.)

Continuous r.v. (cont.)
- remarks
  - for cont. r.v. $P (X = x) = 0, \forall x \in (- \infty, \infty)$
  - $F_{X}$ : continuous
- simple facts
  - $F_{X} (- \infty) = P (X \leq - \infty) = 0$
  - $F_{X} (+ \infty) = P (X \leq + \infty) = 1$
- determining constant in PDF $1 = P (- \infty < X < \infty) = \int + - \infty^{\infty} f_{X} (s) d x$
Problems
- for $X$ a cont. r.v. w/ following PDF: $f_{X} (x) = {c (4 x - 2 x^{2}) 0 0 < x < 2 otherwise$
  - compute value of $c$ $\int_{- \infty}^{\infty} f_{X} (x) d x c = \int_{0}^{2} f_{X} (x) d x = \int_{0}^{2} c (4 x - 2 x^{2}) d x = c (2 x^{2} - \frac{2}{3} x^{3}) ∣_{0}^{2} = c (8 - \frac{16}{3}) = - \frac{8}{3} c = 1 = - \frac{3}{8}$
  - compute value of $P (X > 1)$ $- \frac{3}{8} (2 x^{2} - \frac{2}{3} x^{3}) ∣_{1}^{2} = - \frac{3}{8} ((8 - \frac{16}{3}) - \frac{4}{3}) = \frac{3}{8} \cdot \frac{4}{3} = \frac{1}{2}$
- electrical appliance: function for a random amount of time $T$ $f r (t) = {c e^{- λ t} 0 0 < t < \infty otherwise$
  - exponential distribution
  - CDF: $F_{T} (t) = {1 - e^{- λ t} 0 t \geq 0 otherwise$
  - value of $c$ ? $\int_{0}^{\infty} c e^{- λ t} d t c = - \frac{c e ^{- λ t}}{λ} ∣_{0}^{\infty} = \frac{c}{λ} = 1 = λ$
  - for any $s, t > 0$ , $P (T > s + t ∣ T > s = P (T > t))$ $P (T > t) P (T > s) P (T > s + t) P (T > s + t ∣ T > s) = - \frac{λ e ^{- λ t}}{λ} ∣_{t}^{\infty} = \frac{λ e ^{- λ t}}{λ} = e^{- λ t} = e^{- λ s} = e^{- λ (s + t)} = \frac{e ^{- λ (s + t)}}{e ^{- λ s}} = e^{- λ t} = P (T > t)$
  - property s.t.
    - $P (T > s + t ∣ T > s) = P (T > t)$
      - i.e. memoryless property
- lifetime in hour of a radio tube: given by $f (x) = {0 100/ x^{2} x \leq 100 x > 100$
  - what is the probability that exactly 2 of 5 such tubes: have to be replaced within first 150 hours? $P (T < 150) = \int_{- in f t y}^{150} f (x) d x = \int_{100}^{150} 100/ x^{2} d x = - 100/ x ∣_{100}^{150} = 1 - \frac{2}{3} = \frac{1}{3}$
  - desired property $P (X = 2) = (2 5) P (T < 150)^{2} P (T \geq 150)^{3} = \frac{10 \cdot 2 ^{2}}{3 ^{5}}$

Expectation and Variance of Continuous Random Variables

Expected value and SD
$E (X) var (X) S D (X) = \int_{\infty}^{\infty} x f_{X} (x) d x = \int_{\infty}^{\infty} (x - E (X)) f_{X} (x) d x = E [(X - μ X)^{2}] = var (X)$
- theorem
  - for any real value function $g$
  - ⭐⭐ $E [g (X)] = \int_{- \infty}^{\infty} g (x) f_{X} (s) d x$
  - then same linearity, variance, or variance of linear function
- theorem: tail sum formula
  - for $X$ a nonnegative continuous random variable $E (X) = \int_{0}^{\infty} P (X > x) d x = \int_{0}^{\infty} P (X \geq x) d x$
- proof $\int_{0}^{\infty} P (X > x) d x \int_{0}^{\infty} P (X > x) = \int_{0}^{\infty} \int_{x}^{\infty} f (y) d y d x = \int_{0}^{\infty} (\int_{0}^{y} d x) f (y) d y = \int_{0}^{\infty} y f (y) d y = E [X] = \int_{0}^{\infty} x f (x) d x // order change$
Problems
- find the mean and variance of the random variable $X$ $f (x) = {1/ (b - a) 0 (a \leq x \leq b) otherwise$
  - $X$ : uniform distribution over $[a, b]$ $E (X) = \int_{- \infty}^{\infty} x f (x) d x = \int_{a}^{b} \frac{x}{b - a} d x = \frac{x ^{2}}{2 ( b - a )} ∣_{a}^{b} = \frac{b ^{2} - a ^{2}}{2 ( b - a )} = \frac{a + b}{2}$ $var (X) E (X^{2}) var (X) = E (X^{2}) - (E (X))^{2} = \int_{- \infty}^{\infty} x^{2} f (x) d x = \int_{a}^{b} \frac{x ^{2}}{b - a} d x = \frac{x ^{3}}{3 ( b - a )} ∣_{a}^{b} = \frac{b ^{3} - a ^{3}}{2 ( b - a )} = \frac{a ^{2} + b ^{2} + ab}{3} = \frac{a ^{2} + b ^{2} + ab}{3} - (\frac{a + b}{2})^{2} = \frac{4 a ^{2} + 4 b ^{2} + 4 ab - 3 a ^{2} - 6 ab - 3 b ^{2}}{12} = \frac{a ^{2} - 2 ab + b ^{2}}{12} = \frac{( a - b ) ^{2}}{12}$
- find $E (e^{X})$ w/ $X$ of following pdf $f (x) = {2 x 0 0 \leq x \leq 1 otherwise$ $E (X) = \int_{- \infty}^{\infty} e^{x} f (x) d x = \int_{0}^{1} 2 x e^{x} d x = I BP 2 x e^{x} ∣_{0}^{1} - \int_{0}^{1} e^{x} d x = 2 e - e + 1 = e + 1$
- stick of length 1: broken at random
  - determine expected length of the piece containing point of length $p$ from one end ( $A$ )
  - when $0 \leq p \leq 1$ is fixed
  - let $u$ : break point, then length denoted by $x (u)$ :
    - if $u < p$ then $1 - u$
    - else: $u$
    - $u \sim uniform ([0, 1])$
  - finding: $E [x (u)]$ $E [x (u)] = \int_{- \infty}^{\infty} x (u) f (u) d u = \int_{0}^{1} x (u) f (u) d u = \int_{0}^{p} (1 - u) f (u) d u + \int_{p}^{1} u f (u) d u = (u - \frac{u ^{2}}{2}) ∣_{0}^{p} + (\frac{u ^{2}}{2}) ∣_{p}^{1} = (p - \frac{p ^{2}}{2}) + (\frac{1}{2} - \frac{p ^{2}}{2}) = \frac{1}{2} + p - p^{2}$
- upon meeting: being $s$ minutes early costs you $cs$
  - while being late $s$ minutes cost $k s$
  - suppose: traveling time is a cont. r.v. given by $f$
    - if $x \geq 0$ , given by $f$
    - if $x < 0$ : 0
  - determine: what time to depart in order to minimize cost
  - suppose: you leave $t$ minutes before the appointment
    - step 1: find expected cost $g (t)$
    - step 2: minimize $g (t)$
  - let $C_{t} (X)$ be the cost $C_{t} (X) = {k (x - t) c (t - x) x \geq t = arrive late x < t = arrive early$
  - expected value: $E [C_{t} (x)] = \int_{- \infty}^{\infty} C_{t} (x) f (x) d x = \int_{0}^{\infty} C_{t} (x) f (x) d x = \int_{0}^{t} c (t - x) f (x) d x + \int_{t}^{\infty} k (x - t) f (x) d x = g (t)$
  - minimizing $g (t)$ (finding optimal value, etc.) $g (t) g^{'} (t) = c t \int_{0}^{t} f (x) d x - \int_{0}^{t} x f (x) d x + k \int_{t}^{\infty} x f (x) d x - k t \int_{t}^{\infty} f (x) d x = c t \int_{0}^{t} f (x) d x - \int_{0}^{t} x f (x) d x - k \int_{\infty}^{t} x f (x) d x + k t \int_{\infty}^{t} f (x) d x = c \int_{0}^{t} f (x) d x + c t f (t) - c t f (t) - k t f (t) + k \int_{\infty}^{t} f (x) d x + k t f (t) = c \int_{0}^{t} f (x) d x + k \int_{\infty}^{t} f (x) d x = c \int_{0}^{t} f (x) d x - k \int_{t}^{\infty} f (x) d x$
  - recall that: $F (t) = \int_{- \infty}^{t} f (x) d x = \int_{0}^{t} f (x) d x$
    - and $\int_{t}^{\infty} f (x) d x = \int_{0}^{\infty} f (x) d x - \int_{0}^{t} f (x) d x$
    - $P (X > t) = 1 - P (X \leq t) = 1 - F (t)$
    - $g^{'} (t) = c F (t) - k (1 - F (t)) = (c + k) F (t) - k$
    - $g^{'} (t) = 0 ⟹ (c + k) F (t) = k = 0 ⟹ F (t) = \frac{k}{c + k}$
    - thus $t = F^{- 1} (\frac{k}{c + k})$
  - $g^{''} (t) = [(c + k) F (t) = k]^{'} = (c + k) F^{'} (t) = (c + k) f (t) \geq 0$
    - if $g^{''} (t^{*}) > 0$ $g (t)$ : w/ absolute minimal at $t^{*}$

MATH 2421: Probability