273 lines
7.6 KiB
Text
273 lines
7.6 KiB
Text
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#import "@youwen/zen:0.1.0": *
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#import "@preview/cetz:0.3.2"
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#show: zen.with(
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title: "Homework 5",
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author: "Youwen Wu",
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date: "Winter 2025",
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)
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#show figure: it => {
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pad(y: 10pt, it)
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}
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#set enum(spacing: 2em)
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#let correction = content => {
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set text(fill: red)
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box(stroke: 1pt, inset: 5pt, content)
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}
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#let subproblems = content => {
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set enum(numbering: "a)")
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content
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}
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#rect[
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Initial score: $34/36$
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]
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#rect[
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#set text(fill: red)
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Revised score: $36/36$
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]
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1.
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Define:
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- $S_1 ~ "Bin"(80, p) =$ successful for 55 people
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- $S_2$ = successful for the 2 friends
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- $S_3$ = successful for the other 53 patients
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- $q := 1-p$
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Then we are looking for $P(S_2 | S_1 = 55)$.
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$
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P(S_2 | S_1 = 55) = P(S_2 sect S_1 = 55) / P(S_1 = 55)
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$
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We see that $P(S_2 sect S_1 = 55) = P(S_2) P(S_1 = 55)$.
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$
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P(S_2) &= p^2 \
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P(S_1=55) &= vec(80,55) p^55 q^25 \
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P(S_1) &= vec(78,53) p^53 q^25
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$
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So
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$
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(p^2 vec(78,53) p^53 q^25) / (vec(80,55) p^55 q^25) \
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=(vec(78,53)) / (vec(80,55)) \
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approx 0.4699
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$
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2.
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#subproblems[
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1. We have to make sure it integrates to unity.
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$
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integral_(-infinity)^infinity c e^(-2t) dif t = integral_0^infinity c e^(-2t) dif t &= lr(-1/2 c e^(-2t) |)_(t->infinity) - (lr(-1/2 c e^(-2t) |)_(t=0)) \
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&= 1 / 2 c = 1
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$
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So $c = 2$.
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2. Looking for $P(T > 4)$. This is simply
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$
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1 - P(T <= 4) \
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= 1 - integral_0^4 2e^(-2t) dif t \
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approx 0.000335
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$
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3. Let $X ~ "Bin"(0.00035, 100)$ represent this experiment. We seek $P(X = 6)$.
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$
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P(X=6) = vec(100,6) 0.00035^6 dot (1-0.00035)^94 \
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approx 0
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$
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4. Let $X ~ "Pois"(0.00035 dot 100 = 0.035)$ represent this experiment. We seek $P(X = 6)$.
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$
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P(X=6) = e^(-0.035) (0.035^6) / 6! approx 0
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$
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5.
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Yes, $p < 0.02$ and $n >= 100$.
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]
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3. #subproblems[
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1. Check the left and right limits at infinity: $F(-infinity) = 0$, $F(infinity) = 1$.
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$F$ can never take a value greater than 1 or less than 0.
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$dif/(dif x) F(x) = e^(-x)$ which is positive for $x >= 0$ so it is
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monotonically increasing.
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Therefore it is a valid CMF (i).
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For (ii), we just differentiate.
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$ f(x) = dif / (dif x) F(x) = e^(-x) $
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2. Check the left and right limits at infinity: $F(-infinity) = 0$, $F(infinity) = 0$.
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So immediately we see it is not valid.
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]
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4. #subproblems[
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1. We are looking for $P(X <= x)$, the CDF. Immediately notice that the state
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space of $X$ is $0 <= X <= 0.5$. When asking for $P(X <= x)$, we are essentially asking the probability that our uniform point lies in this red region:
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#figure(
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align(
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center,
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cetz.canvas({
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import cetz.draw: *
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rect((0, 0), (5, 5), fill: red)
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rect((1, 1), (4, 4), fill: white)
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}),
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),
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caption: "point lies in the inset border",
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)
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But it seems much easier to find if the point lies in the center region
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instead, which is $P(X >= x)$. We want the probability that $X$ is _at least_
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$x$ distance away from the sides, which is basically asking if it lies within
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a smaller square inset by $x$ from the unit square.
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#figure(
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align(
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center,
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cetz.canvas({
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import cetz.draw: *
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rect((0, 0), (5, 5), fill: white)
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rect((1, 1), (4, 4), fill: red)
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}),
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),
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caption: "point lies in the center",
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)
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Intuitively, trying to add up the values of many individual points sounds a
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lot like an area!
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What is the area of this square? The inset is $x$, so its side lengths are $1
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- 2x$, and the area is therefore $4x^2 - 4x$. Then $P(X >= x) = "area of red square"/"area of unit square"$, and the unit square has area 1, so $P(X >= x) = 4x^2-4x$.
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Then the CDF, $P(X <= x)$ is just
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$ 1- P(X >=x) = 4x - 4x^2 $
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2. The PDF is just the derivative of the CDF so
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$
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dif / (dif x) 4x - 4x^2 = 4 - 8x
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$
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#correction[
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$-2$ points. We forgot to ensure the CDF is actually valid.
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We need to make sure that the limits at infinity are 0 and 1. So
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$
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F_x = cases(4x-4x^2"," &"when" 0 <= x < 0.5, 0"," &"when" x < 0, 1"," &"when" x >= 0.5)
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$
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Also we should check that $F_x$ is monotone increasing (true) and always
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lies in $[0,1]$ (true).
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This means that the PDF is
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$
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f_x = cases(4 - 8x"," &"when" 0 <= x < 0.5, 0"," &"otherwise")
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$
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]
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]
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5. If we assume that each typo occurs independently and with a very small fixed
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rate $p$, and each page has the same amount of words (with $>= 100$ words per
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page), then this can be characterized by a Poisson distribution. To find
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$lambda$, simply recall that the expected value is the mean and equal to
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$lambda$ for the Poisson. Then the number of typos on each page can be
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characterized by a Poisson
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distribution $X ~ "Pois"(6)$ and we seek $P(X >= 4)$.
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$
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P(X >= 4) = 1 - P(X < 4) = 1 - sum_(k=0)^3 e^(-6) (6^k) / k! approx 0.8488
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$
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6. #subproblems[
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1. We need to check it integrates to unity.
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$
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integral_(-infinity)^infinity f_X (x) dif x = integral_0^infinity 3e^(-3x) dif x = 1
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$
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2. $
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PP(-1 < X < 1) = integral_(-1)^1 3e^(-3x) dif x approx 0.9502
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$
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3. $
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PP(X < 5) = integral_(-infinity)^5 f_X dif x approx 1
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$
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4. $
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PP(2 < X < 4 | X < 5) = PP(2 < X < 4) / PP(X < 5) = (integral_2^4 f_X (x) dif x) / (integral_0^5 f_X (x) dif x) approx 0.002473
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$
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5.
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$
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F_X (x) = cases(1 - e^(-3x)"," &"when" x > 0, 0"," &"otherwise")
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$
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]
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7. $
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F_X (x) = integral_(-infinity)^x f_X (t) dif t \
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1 - F_X (x) = integral_x^infinity f_X (t) dif t \
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integral_0^infinity 1 - F_x (x) dif x = integral_0^infinity integral_0^t f_X (t) dif x dif t \
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= integral_0^infinity t f_X (t) dif t
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$
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8. #subproblems[
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1. #figure(
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cetz.canvas({
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import cetz.draw: *
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set-style(stroke: 0.4pt)
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grid((-3, -3), (3, 3), step: 1.5, stroke: gray + 0.2pt)
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line((-3, 0), (3, 0))
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line((0, -3), (0, 3))
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rect((0, 0), (1.5, 1.5), fill: gray)
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line((0, 0), (1.5, 0.75))
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line((0, 0), (1.5, 1.5))
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line((0, 0), (0.75, 1.5))
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}),
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)
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Note that the slope of a line through the origin is determined entirely by
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any point it passes through. Therefore given a line slope $s$, any point
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$(x,y)$ on this line satisfies slope $s = y/x$. We are for $P(S < s)$, but
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notice that every slope less than $s$ is also given by all of the points in
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the unit square on the line it defines. This means that we can reframe our
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question as asking to choose a point uniformly in the unit square, and
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finding the probability that it is located in the region underneath the line!
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The probability is given by the ratio of the area of all under the line
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divided by the area of the square. But since the area of the square is 1,
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it's simply just the area under the line.
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When the slope is less than 1, this probability is just the area of the triangle
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formed by the edges of the square and the line through the origin. When the
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slope is greater than 1, it's equal to 1 minus the area of the triangle.
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(when the slope is $1$ the area is always $1/2$).
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So the CDF is
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$
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F_S (s) = P(S <= s) = cases(1/2 s"," &"when" 0 < s <= 1, 1 - 1/(2s)"," &"when" < 1 < infinity, 0"," &"otherwise")
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$
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2. The PDF is just the derivative
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$
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f_S (s) = cases(1/2 "," &"when" 0 < s <= 1, 1/(2s^2)"," &"when" < 1 < infinity, 0"," &"otherwise")
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$
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]
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