auto-update(nvim): 2025-02-03 15:15:35
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@ -365,9 +365,9 @@ This is mostly a formal manipulation to derive the obviously true proposition fr
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$
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]
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#proposition[
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#proposition("Inclusion-exclusion principle")[
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$ P(A union B) = P(A) + P(B) - P(A sect B) $
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]
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]<inclusion-exclusion>
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#proof[
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$
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@ -1143,3 +1143,269 @@ exactly one sequence that gives us success.
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(5 / 6)^7
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$
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]
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= Lecture #datetime(day: 3, month: 2, year: 2025).display()
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== CDFs, PMFs, PDFs
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Properties of a CDF:
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Any CDF $F(x) = P(X <= x)$ satisfies
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1. $F(-infinity) = 0$, $F(infinity) = 1$
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2. $F(x)$ is non-decreasing in $x$ (monotonically increasing)
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$ s < t => F(s) <= F(t) $
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3. $P(a < X <= b) = P(X <= b) - P(X <= a) = F(b) - F(a)$
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#example[
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Let $X$ be a continuous random variable with density (pdf)
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$
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f(x) = cases(
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c x^2 &"for" 0 < x < 2,
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0 &"otherwise"
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)
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$
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1. What is $c$?
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$c$ is such that
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$
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1 = integral^infinity_(-infinity) f(x) dif x = integral_0^2 c x^2 dif x
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$
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2. Find the probability that $X$ is between 1 and 1.4.
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Integrate the curve between 1 and 1.4.
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$
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integral_1^1.4 3 / 8 x^2 dif x = (x^3 / 8) |_1^1.4 \
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= 0.218
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$
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This is the probability that $X$ lies between 1 and 1.4.
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3. Find the probability that $X$ is between 1 and 3.
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Idea: integrate between 1 and 3, be careful after 2.
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$ integral^2_1 3 / 8 x^2 dif x + integral_2^3 0 dif x = $
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4. What is the CDF for $P(X <= x)$? Integrate the curve to $x$.
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$
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F(x) = P(X <= x) = integral_(-infinity)^x f(t) dif t \
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= integral_0^x 3 / 8 t^2 dif t \
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= x^3 / 8
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$
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Important: include the range!
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$
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F(x) = cases(
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0 &"for" x <= 0,
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x^3/8 &"for" 0 < x < 2,
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1 &"for" x >= 2
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)
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$
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5. Find a point $a$ such that you integrate up to the point to find exactly $1/2$
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the area.
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We want to find $1/2 = P(X <= a)$.
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$ 1 / 2 = P(X <= a) = F(a) = a^3 / 8 => a = root(3, 4) $
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]
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== The (continuous) uniform distribution
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The most simple and the best of the named distributions!
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#definition[
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Let $[a,b]$ be a bounded interval on the real line. A random variable $X$ has the uniform distribution on the interval $[a,b]$ if $X$ has the density function
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$
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f(x) = cases(
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1/(b-a) &"for" x in [a,b],
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0 &"for" x in.not [a,b]
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)
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$
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Abbreviate this by $X ~ "Unif" [a,b]$.
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]<continuous-uniform>
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The graph of $"Unif" [a,b]$ is a constant line at height $1/(b-a)$ defined
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across $[a,b]$. The integral is just the area of a rectangle, and we can check
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it is 1.
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#fact[
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For $X ~ "Unif" [a,b]$, its cumulative distribution function (CDF) is given by:
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$
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F_x (x) = cases(
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0 &"for" x < a,
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(x-a)/(b-a) &"for" x in [a,b],
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1 &"for" x > b
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)
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$
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]
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#fact[
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If $X ~ "Unif" [a,b]$, and $[c,d] subset [a,b]$, then
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$
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P(c <= X <= d) = integral_c^d 1 / (b-a) dif x = (d-c) / (b-a)
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$
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]
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#example[
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Let $Y$ be a uniform random variable on $[-2,5]$. Find the probability that its
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absolute value is at least 1.
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$Y$ takes values in the interval $[-2,5]$, so the absolute value is at least 1 iff. $Y in [-2,1] union [1,5]$.
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The density function of $Y$ is $f(x) = 1/(5- (-2)) = 1/7$ on $[-2,5]$ and 0 everywhere else.
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So,
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$
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P(|Y| >= 1) &= P(Y in [-2,-1] union [1,5]) \
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&= P(-2 <= Y <= -1) + P(1 <= Y <= 5) \
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&= 5 / 7
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$
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]
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== The exponential distribution
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The geometric distribution can be viewed as modeling waiting times, in a discrete setting, i.e. we wait for $n - 1$ failures to arrive at the $n^"th"$ success.
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The exponential distribution is the continuous analogue to the geometric
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distribution, in that we often use it to model waiting times in the continuous
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sense. For example, the first custom to enter the barber shop.
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#definition[
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Let $0 < lambda < infinity$. A random variable $X$ has the exponential distribution with parameter $lambda$ if $X$ has PDF
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$
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f(x) = cases(
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lambda e^(-lambda x) &"for" x >= 0,
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0 &"for" x < 0
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)
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$
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Abbreviate this by $X ~ "Exp"(lambda)$, the exponential distribution with rate $lambda$.
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The CDF of the $"Exp"(lambda)$ distribution is given by:
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$
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F(t) + cases(
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0 &"if" t <0,
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1 - e^(-lambda t) &"if" t>= 0
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)
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$
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]
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#example[
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Suppose the length of a phone call, in minutes, is well modeled by an exponential random variable with a rate $lambda = 1/10$.
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1. What is the probability that a call takes more than 8 minutes?
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2. What is the probability that a call takes between 8 and 22 minutes?
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Let $X$ be the length of the phone call, so that $X ~ "Exp"(1/10)$. Then we can find the desired probability by:
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$
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P(X > 8) &= 1 - P(X <= 8) \
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&= 1 - F_x (8) \
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&= 1 - (1 - e^(-(1 / 10) dot 8)) \
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&= e^(-8 / 10) approx 0.4493
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$
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Now to find $P(8 < X < 22)$, we can take the difference in CDFs:
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$
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&P(X > 8) - P(X >= 22) \
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&= e^(-8 / 10) - e^(-22 / 10) \
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&approx 0.3385
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$
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]
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#fact("Memoryless property of the exponential distribution")[
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Suppose that $X ~ "Exp"(lambda)$. Then for any $s,t > 0$, we have
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$
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P(X > t + s | X > t) = P(X > s)
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$
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]<memoryless>
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This is like saying if I've been waiting 5 minutes and then 3 minutes for the
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bus, what is the probability that I'm gonna wait more than 5 + 3 minutes, given
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that I've already waited 5 minutes? And that's precisely equal to just the
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probability I'm gonna wait more than 3 minutes.
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#proof[
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$
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P(X > t + s | X > t) = (P(X > t + s sect X > t)) / (P(X > t)) \
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= P(X > t + s) / P(X > t)
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= e^(-lambda (t+ s)) / (e^(-lambda t)) = e^(-lambda s) \
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equiv P(X > s)
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$
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]
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== Gamma distribution
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#definition[
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Let $r, lambda > 0$. A random variable $X$ has the *gamma distribution* with parameters $(r, lambda)$ if $X$ is nonnegative and has probability density function
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$
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f(x) = cases(
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(lambda^r x^(r-2))/(Gamma(r)) e^(-lambda x) &"for" x >= 0,
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0 &"for" x < 0
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)
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$
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Abbreviate this by $X ~ "Gamma"(r, lambda)$.
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]
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The gamma function $Gamma(r)$ generalizes the factorial function and is defined as
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$
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Gamma(r) = integral_0^infinity x^(r-1) e^(-x) dif x, "for" r > 0
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$
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Special case: $Gamma(n) = (n - 1)!$ if $n in ZZ^+$.
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#remark[
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The $"Exp"(lambda)$ distribution is a special case of the gamma distribution,
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with parameter $r = 1$.
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]
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== The normal (Gaussian) distribution
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#definition[
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A random variable $ZZ$ has the *standard normal distribution* if $Z$ has
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density function
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$
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phi(x) = 1 / sqrt(2 pi) e^(-x^2 / 2)
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$
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on the real line. Abbreviate this by $Z ~ N(0,1)$.
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]<normal-dist>
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#fact("CDF of a standard normal random variable")[
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Let $Z~N(0,1)$ be normally distributed. Then its CDF is given by
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$
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Phi(x) = integral_(-infinity)^x phi(s) dif s = integral_(-infinity)^x 1 / sqrt(2 pi) e^(-(-s^2) / 2) dif s
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$
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]
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The normal distribution is so important, instead of the standard $f_Z(x)$ and
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$F_z(x)$, we use the special $phi(x)$ and $Phi(x)$.
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#fact[
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$
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integral_(-infinity)^infinity e^(-s^2 / 2) dif s = sqrt(2 pi)
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$
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No closed form of the standard normal CDF $Phi$ exists, so we are left to either:
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- approximate
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- use technology (calculator)
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- use the standard normal probability table in the textbook
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]
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