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An introduction to continuous distributions, focusing on the uniform, exponential, and gamma distributions. It covers the concepts of probability density functions (pdf) and cumulative distribution functions (cdf), and includes examples and formulas for calculating probabilities and percentiles. The document also discusses the relationship between pdf and cdf, and provides an explanation of how these distributions are used to model real-world phenomena.
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1.8-1.9: Continuous Random Variables
1.10.1: Uniform Distribution (Continuous)
1.10.4-5 Exponential and Gamma Distributions: Distance between crossovers
Prof. Tesler
Math 283 Fall 2015
Pick a real number x between 20 and 30 with all real values in [ 20 , 30 ] equally likely. Sample space: S = [ 20 , 30 ] Number of outcomes: |S| = ∞ Probability of each outcome: P(X = x) = (^) ∞^1 = 0 Yet, P(X 6 21. 5 ) = 15 %
We could regard the sample space as [ 20 , 30 ], or as all reals.
fX(x) =
1 / 10 for 20 6 x 6 30 ; 0 otherwise.
x
f((X
x))
0 10 20 30 40
−∞
0 dx +
20
dx = 0 +
x 10
=
x
f((X
x))
0 10 20 30 40
The Cumulative Distribution Function (cdf) of a random variable X is FX(x) = P(X 6 x) For a continuous random variable, FX(x) = P(X 6 x) =
∫x −∞ fX(t)^ dt^ and^ fX(x) =^ FX
′(x)
The integral cannot have “x” as the name of the variable in both of FX(x) and fX(x) because one is the upper limit of the integral and the other is the integration variable. So we use two variables x, t. We can either write FX(x) = P(X 6 x) =
∫ (^) x
−∞
fX(t) dt or FX(t) = P(X 6 t) =
∫ (^) t
−∞
fX(x) dx
Probability density function
x
f((X
x))
0 10 20 30 40
fX(x) =
. 1 if 20 6 x 6 30 ; 0 otherwise. It’s discontinuous at x = 20 and 30. PDF is derivative of CDF: fX(x) = FX^ ′(x)
Cumulative distribution function
x
FX
((x))
0 10 20 30 40
0
1
F X(x) =
0 if x < 20 ; (x − 20 )/ 10 if 20 6 x 6 30 ; 1 if x > 30. CDF is integral of PDF: FX(x) =
∫ (^) x
−∞
fX(t) dt
0 1 2 3
0
Probability density function
r
density f
(r)R
fR(r) =
2 r/ 9 if 0 6 r < 3 ; 0 if r 6 0 or r > 3 It’s discontinuous at r = 3. PDF is derivative of CDF: fR(r) = FR^ ′(r)
0 1 2 3
0
1
Cumulative distribution function
r
FR
(r)
FR(r) =
0 if r < 0 ; r^2 / 9 if 0 6 r 6 3 ; 1 if r > 3. CDF is integral of PDF: FR(r) =
∫ (^) r
−∞
fR(t) dt
(^00 1 2 )
1
Cumulative distribution function
r
FR
(r)
(^0)! 1 0 1 2
1
Cumulative distribution function
x
FX
(x)
In a continuous distribution: The probability of an individual point is 0 : P(R = r) = 0. So, P(R 6 r) = P(R < r), i.e., FR(r) = FR(r−). The CDF is continuous. (In a discrete distribution, the CDF is discontinuous due to jumps at the points with nonzero probability.) P(a < R < b)= P(a 6 R < b) = P(a < R 6 b) = P(a 6 R 6 b) = FR(b) − FR(a)
The Cumulative Distribution Function (cdf) of a random variable X is FX(x) = P(X 6 x)
FX(x) =
∫x −∞ fX(t)^ dt Weakly increasing. Varies smoothly from 0 to 1 as x varies from −∞ to ∞. To get the pdf from the cdf, use fX(x) = FX^ ′(x).
FX(x) =
t 6 x PX(t) Weakly increasing. Stair-steps from 0 to 1 as x goes from −∞ to ∞. The cdf jumps where PX(x) , 0 and is constant in-between. To get the pdf from the cdf, use PX(x) = FX(x) − FX(x−) (which is positive at the jumps, 0 otherwise).
Replace sums by integrals. It’s the same definitions in terms of “E(·)”:
μ = E(X) =
−∞
x · fX(x) dx
E(g(X)) =
−∞
g(x) fX(x) dx
σ^2 = Var(X) = E((X − μ)^2 ) = E(X^2 ) − (E(X))^2
μ = E(X) =
∫ (^) b
a
x ·
b − a
dx =
x^2 / 2 b − a
b
x=a
(b^2 − a^2 )/ 2 b − a
b + a 2
E(X^2 ) =
∫ (^) b
a
x^2 ·
b − a
dx =
x^3 / 3 b − a
b
x=a
(b^3 − a^3 )/ 3 b − a
b^2 + ab + a^2 3
σ^2 = Var(X) = E(X^2 ) − (E(X))^2 =
b^2 + ab + a^2 3
b + a 2
(b − a)^2 12 σ = SD(X) = (b − a)/
How far is it from the start of a chromosome to the first crossover? How far is it from one crossover to the next? Let D be the random variable giving either of those. It is a real number > 0 , with the exponential distribution
fD(d) =
λ e−λ^ d^ if d > 0 ; 0 if d < 0. where crossovers happen at a rate λ = 1 M−^1 = 0. 01 cM−^1. General case Crossovers Mean E(D) = 1 /λ = 100 cM = 1 M Variance Var(D) = 1 /λ^2 = 10000 cM^2 = 1 M^2 Standard Dev. SD(D) = 1 /λ = 100 cM = 1 M
In general, if events occur on the real number line x > 0 in such a way that the expected number of events in all intervals [x, x + d] is λ d (for x > 0 ), then the exponential distribution with parameter λ models the time/distance/etc. until the first event.
It also models the time/distance/etc. between consecutive events.
Chromosomes are finite; to make this model work, treat “there is no next crossover” as though there is one but it happens somewhere past the end of the chromosome.
Let d > 0 be any real number. Let N(d) be the # of crossovers that occur in the interval [ 0 , d].
0 d
D>d N(d)= D<d N(d)= D<d N(d)=
^
If N(d) = 0 then there are no crossovers in [ 0 , d], so D > d. If D > d then the first crossover is after d so N(d) = 0. Thus, D > d is equivalent to N(d) = 0. P(D > d) = P(N(d) = 0 ) = e−λ^ d(λ d)^0 / 0! = e−λ^ d since N(d) has a Poisson distribution with parameter λ d. The cdf of D is FD(d) = P(D 6 d) = 1 − P(D > d) =
1 − e−λ^ d^ if d > 0 ; 0 if d < 0. Differentiating the cdf gives pdf fD(d) = FD^ ′(d) = λ e−λ^ d^ (if d > 0 ).
How far is it from the start of a chromosome until the rth crossover, for some choice of r = 1 , 2 , 3 ,.. .? Let Dr be a random variable giving this distance. It has the gamma distribution with pdf
fDr(d) =
{ (^) λr (r− 1 )! d
r− (^1) e−λ d (^) if d > 0 ;
0 if d < 0.
Mean E(Dr) = r/λ Variance Var(Dr) = r/λ^2 Standard deviation SD(Dr) =
r/λ The gamma distribution for r = 1 is the same as the exponential distribution. The sum of r i.i.d. exponential variables, Dr = X 1 + X 2 + · · · + Xr, each with rate λ, gives the gamma distribution.
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