L11 Continuous Distributions (WMS 4.1-3)

Continuous random variables
- Infinite domain, e.g. sleep hours x\in\lbrack 6,9\rbrack
- Philosophical view: continuous functions conveniently approximate discrete world, or world is truly infinite but measurement is imprecise
Probability density function (pdf) f(x)
- Measures relative likelihood of individual x values
- Individual x values occur with zero probability (and \int_7^8f(x) > 1 is possible); to find probabilities, must take definite integral P(7 < X < 8) = f(x)\text{dx}
- Density must be non-negative and integrate to one over domain (just like probabilities sum to one)
- Example f(x) = k(- x^{2} + 16x - 60); 6\leq x\leq 9
  1. Not directly from (finite) data; maybe from calibrated theory
  2. Find k
    1. 1 = \int_6^9f(x)dx = k\lbrack -\frac{1}{3}x^{3} + 8x^{2} - 60x\rbrack_{6}^{9} = k\lbrack (- 243 + 648 - 540) -(- 72 + 288 - 360)\rbrack = 9k requires that k =\frac{1}{9}
    2. That is, f(x) = -\frac{1}{9}x^{2} +\frac{16}{9}x -\frac{60}{9}; 6\leq x\leq 9
  3. Mode solves f^{'}(x) = -\frac{2}{9}kx +\frac{16}{9}k = 0; solution at x = 8
    1. Note: if f^{'}(x) everywhere positive/negative then maximum is at highest/lowest x in range
    2. Note: second-order condition f^{''}(x) = -\frac{2}{9}k\leq 0 satisfied as long as k\geq 0
  4. Probabilities: P(7\leq x\leq 8) =\frac{1}{9}(- x^{2} + 16x - 60)dx =\ldots =\frac{11}{27}\approx 0.4
Cumulative distribution function (cdf) F(x)
- $F(x) = P(X\leq x) =\frac{1}{9}(- {\widetilde{x}}^{2} + 16\widetilde{x} - 60)d\widetilde{x}$ =\lbrack -\frac{1}{27}{\widetilde{x}}^{3} +\frac{8}{9}{\widetilde{x}}^{2} -\frac{20}{3}\widetilde{x}\rbrack_{\widetilde{x} = 6}^{\widetilde{x} = x} = -\frac{1}{27}x^{3} +\frac{8}{9}x^{2} -\frac{20}{3}x + 16 (This assumes 6\leq x\leq 9; if x < 6 then F(x) = 0 and if $x > 9$then F(x) = 1)
- Percentiles Median F(x) = -\frac{1}{27}x^{3} +\frac{8}{9}x^{2} -\frac{20}{3}x + 16 =\frac{1}{2}; solving by computer, x\approx 7.8 75^{th} percentile F(x) = -\frac{1}{27}x^{3} +\frac{8}{9}x^{2} -\frac{20}{3}x + 16 = .75\Rightarrow x\approx 8.4 95^{th} percentile F(x) = -\frac{1}{27}x^{3} +\frac{8}{9}x^{2} -\frac{20}{3}x + 16 = .90\Rightarrow x\approx 8.7
- Easier probabilities $P(7\leq X\leq 8) = F(8) - F(7)$ =(-\frac{1}{27}8^{3} +\frac{8}{9}8^{2} -\frac{20}{3}8 + 16) -(-\frac{1}{27}7^{3} +\frac{8}{9}7^{2} -\frac{20}{3}7 + 16) =\frac{11}{27}\approx 0.4
- From cdf, get pdf f(x) = F^{'}(x) = -\frac{1}{9}x^{2} +\frac{16}{9}x -\frac{60}{9}; 6\leq x\leq 9, else f(x) = 0
Moments
- Mean \mu = E(X) =\int\text{xf}(x)\text{dx} (just like E(X) =\sum xP(x)) = \int_6^9x\frac{1}{9}(- x^{2} + 16x - 60)\text{dx} =\int_6^9\frac{1}{9}(- x^{3} + 16x^{2} - 60x)dx =\ldots =\frac{31}{4}\approx 7.75

Standard deviation
1. $E(X^{2}) = \int_6^9x^{2}f(x)\text{dx}$ = \int_6^9x^{2}\frac{1}{9}(- x^{2} + 16x - 60)dx =\int_6^9\frac{1}{9}(- x^{4} + 16x^{3} - 60x^{2})dx =\ldots =\frac{303}{5}

ii. V(X) = E(X^{2}) -\mu^{2} =\frac{303}{5} -(\frac{31}{4})^{2} =\frac{43}{80}

iii. \sigma_{X} =\sqrt{\frac{43}{80}}\approx 0.73

Note: algebra tricks still work (e.g. lost wages while sleeping)
1. E(\$ 20X) =\$ 20E(X) =\$ 20\cdot 7.75 =\$ 155
2. V(20X) = 20^{2}V(X)

Practice describing steps to classmate: Warehouse stock (as fraction of capacity) f(x) = - 2x^{2} + kx +\frac{1}{6};0\leq x\leq 1
- Find k = 3
- mode =\frac{3}{4}
- Draw and interpret pdf (upside-down parabola; warehouse full more often than empty)
- Find cdf F(x) = -\frac{2}{3}x^{3} +\frac{3}{2}x^{2} +\frac{1}{6}x;0\leq x\leq 1
- Find f(x) from F(x)
- P(\frac{1}{2}\leq X\leq\frac{3}{4}) =\frac{5}{16}
- median \approx .6, 75^th^ percentile \approx .8
- mean \mu\approx 0.58
- standard deviation \sigma\approx 0.26
- Insurance payout \pi =\$ 1,000,000X +\$ 100,000
  1. E(\pi) =\$ 1,000,000\mu +\$ 100,000 =\$ 680,000 $\sigma_{\pi} = \sqrt{V($1,000,000X+$100,000)} =$ 1,000,000\sigma_{x} =$ 260,000$

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L11 Continuous Distributions (WMS 4.1-3)

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