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Probability distributions: Who cares & why?
Arnab Chakraborty
Indian Statistical Institute
Nov 12, 2017
“Snakes and Ladders” ludo
“Snakes and Ladders” ludo
“Snakes and Ladders” ludo
A strange ludo
1.Xnew = 0.8Xold + 0.1Ynew = 0.8Yold + 0.04
2.Xnew = 0.5Xold + 0.25Ynew = 0.5Yold + 0.4
3.Xnew = 0.355Xold − 0.355Yold +0.266Ynew = 0.355Xold +0.355Yold +0.078
4.Xnew = 0.355Xold + 0.355Yold + 0.378Ynew = −0.355Xold + 0.355Yold + 0.434
A strange ludo
1.Xnew = 0.8Xold + 0.1Ynew = 0.8Yold + 0.04
2.Xnew = 0.5Xold + 0.25Ynew = 0.5Yold + 0.4
3.Xnew = 0.355Xold − 0.355Yold +0.266Ynew = 0.355Xold +0.355Yold +0.078
4.Xnew = 0.355Xold + 0.355Yold + 0.378Ynew = −0.355Xold + 0.355Yold + 0.434
A strange ludo
1.Xnew = 0.8Xold + 0.1Ynew = 0.8Yold + 0.04
2.Xnew = 0.5Xold + 0.25Ynew = 0.5Yold + 0.4
3.Xnew = 0.355Xold − 0.355Yold +0.266Ynew = 0.355Xold +0.355Yold +0.078
4.Xnew = 0.355Xold + 0.355Yold + 0.378Ynew = −0.355Xold + 0.355Yold + 0.434
A strange ludo
1.Xnew = 0.8Xold + 0.1Ynew = 0.8Yold + 0.04
2.Xnew = 0.5Xold + 0.25Ynew = 0.5Yold + 0.4
3.Xnew = 0.355Xold − 0.355Yold +0.266Ynew = 0.355Xold +0.355Yold +0.078
4.Xnew = 0.355Xold + 0.355Yold + 0.378Ynew = −0.355Xold + 0.355Yold + 0.434
A strange thing!
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0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
100
A strange thing!
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0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
500
A strange thing!
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
1000
A strange thing!
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
10000
A strange thing!
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
100000
Try it out yourself!
https://arnab-chakraborty.shinyapps.io/shny/
Another strange thing!
−2 0 2
02
46
810
100000
Statistical regularity
Regularity in randomness!
I Not always
I Only when we have “lots of randomness”
Natural phenomena:
I Leaves: very similar but not same
I Finger prints
Statistical regularity
Regularity in randomness!
I Not always
I Only when we have “lots of randomness”
Natural phenomena:
I Leaves: very similar but not same
I Finger prints
Why care?
1. Understanding: Probability
2. Using: Statistics
Why care?
1. Understanding: Probability
2. Using: Statistics
Why care?
1. Understanding: Probability
2. Using: Statistics
Histogram
Histogram
Histogram
Histogram
Probability density function
0.0 0.2 0.4 0.6 0.8 1.0
01
23
Probability density function
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.5
1.0
1.5
2.0
2.5
Probability density function
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.5
1.0
1.5
2.0
2.5
Probability density function
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.5
1.0
1.5
2.0
Probability density function
Probabilitydensity function
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.5
1.0
1.5
2.0
Probability density function
Probabilitydensity function
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.5
1.0
1.5
2.0
Probability density function
Probabilitydensity function
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.5
1.0
1.5
2.0
All continuous randomvariables have PDFs.
Probability density function
Probability density function
Probability density function
Probability density function
Probability density function
Probability density function
Probability density function
All continuous randomvariable pairs have jointPDFs.
Molecules
Molecules
A single molecule
A single molecule
A single molecule
A single molecule
A single molecule
A single molecule
Isotropy
x , y , z are continuous random variables and so havePDFs.
In case of ”no flow”
I They have the same density (call it f (·))I They are independent.
I So joint density of x , y , z is f (x)f (y)f (z).
I f (x)f (y)f (z) does not depends only on the length of(x , y , z) and not on the direction.
Isotropy
x , y , z are continuous random variables and so havePDFs.
In case of ”no flow”
I They have the same density (call it f (·))
I They are independent.
I So joint density of x , y , z is f (x)f (y)f (z).
I f (x)f (y)f (z) does not depends only on the length of(x , y , z) and not on the direction.
Isotropy
x , y , z are continuous random variables and so havePDFs.
In case of ”no flow”
I They have the same density (call it f (·))I They are independent.
I So joint density of x , y , z is f (x)f (y)f (z).
I f (x)f (y)f (z) does not depends only on the length of(x , y , z) and not on the direction.
Isotropy
x , y , z are continuous random variables and so havePDFs.
In case of ”no flow”
I They have the same density (call it f (·))I They are independent.
I So joint density of x , y , z is f (x)f (y)f (z).
I f (x)f (y)f (z) does not depends only on the length of(x , y , z) and not on the direction.
Isotropy
x , y , z are continuous random variables and so havePDFs.
In case of ”no flow”
I They have the same density (call it f (·))I They are independent.
I So joint density of x , y , z is f (x)f (y)f (z).
I f (x)f (y)f (z) does not depends only on the length of(x , y , z) and not on the direction.
Isotropy
Isotropy
Isotropy
Mathematically...
f (x)f (y)f (z) = g(x2 + y 2 + z2)
f ′(x)f (y)f (z) = 2xg ′(x2 + y 2 + z2)
f (x)f ′(y)f (z) = 2yg ′(x2 + y 2 + z2)
f (x)f (y)f ′(z) = 2zg ′(x2 + y 2 + z2)
g ′(x2+y 2+z2) =f ′(x)f (y)f (z)
2x=
f (x)f ′(y)f (z)
2y=
f (x)f (y)f ′(z)
2z.
f ′(x)
xf (x)=
f ′(y)
yf (y)=
f ′(z)
zf (z)
= k , say
.
Mathematically...
f (x)f (y)f (z) = g(x2 + y 2 + z2)
f ′(x)f (y)f (z) = 2xg ′(x2 + y 2 + z2)
f (x)f ′(y)f (z) = 2yg ′(x2 + y 2 + z2)
f (x)f (y)f ′(z) = 2zg ′(x2 + y 2 + z2)
g ′(x2+y 2+z2) =f ′(x)f (y)f (z)
2x=
f (x)f ′(y)f (z)
2y=
f (x)f (y)f ′(z)
2z.
f ′(x)
xf (x)=
f ′(y)
yf (y)=
f ′(z)
zf (z)
= k , say
.
Mathematically...
f (x)f (y)f (z) = g(x2 + y 2 + z2)
f ′(x)f (y)f (z) = 2xg ′(x2 + y 2 + z2)
f (x)f ′(y)f (z) = 2yg ′(x2 + y 2 + z2)
f (x)f (y)f ′(z) = 2zg ′(x2 + y 2 + z2)
g ′(x2+y 2+z2) =f ′(x)f (y)f (z)
2x=
f (x)f ′(y)f (z)
2y=
f (x)f (y)f ′(z)
2z.
f ′(x)
xf (x)=
f ′(y)
yf (y)=
f ′(z)
zf (z)
= k , say
.
Mathematically...
f (x)f (y)f (z) = g(x2 + y 2 + z2)
f ′(x)f (y)f (z) = 2xg ′(x2 + y 2 + z2)
f (x)f ′(y)f (z) = 2yg ′(x2 + y 2 + z2)
f (x)f (y)f ′(z) = 2zg ′(x2 + y 2 + z2)
g ′(x2+y 2+z2) =f ′(x)f (y)f (z)
2x=
f (x)f ′(y)f (z)
2y=
f (x)f (y)f ′(z)
2z.
f ′(x)
xf (x)=
f ′(y)
yf (y)=
f ′(z)
zf (z)
= k , say
.
Mathematically...
f (x)f (y)f (z) = g(x2 + y 2 + z2)
f ′(x)f (y)f (z) = 2xg ′(x2 + y 2 + z2)
f (x)f ′(y)f (z) = 2yg ′(x2 + y 2 + z2)
f (x)f (y)f ′(z) = 2zg ′(x2 + y 2 + z2)
g ′(x2+y 2+z2) =f ′(x)f (y)f (z)
2x=
f (x)f ′(y)f (z)
2y=
f (x)f (y)f ′(z)
2z.
f ′(x)
xf (x)=
f ′(y)
yf (y)=
f ′(z)
zf (z)= k , say.
Solving
df
dx= kxf .
df
f= kxdx .∫
df
f= k
∫xdx .
log f =kx2
2+ const.
f = const × ekx2
2 .
Maxwell / Gaussian distribution.
Solving
df
dx= kxf .
df
f= kxdx .
∫df
f= k
∫xdx .
log f =kx2
2+ const.
f = const × ekx2
2 .
Maxwell / Gaussian distribution.
Solving
df
dx= kxf .
df
f= kxdx .∫
df
f= k
∫xdx .
log f =kx2
2+ const.
f = const × ekx2
2 .
Maxwell / Gaussian distribution.
Solving
df
dx= kxf .
df
f= kxdx .∫
df
f= k
∫xdx .
log f =kx2
2+ const.
f = const × ekx2
2 .
Maxwell / Gaussian distribution.
Solving
df
dx= kxf .
df
f= kxdx .∫
df
f= k
∫xdx .
log f =kx2
2+ const.
f = const × ekx2
2 .
Maxwell / Gaussian distribution.
Solving
df
dx= kxf .
df
f= kxdx .∫
df
f= k
∫xdx .
log f =kx2
2+ const.
f = const × ekx2
2 .
Maxwell / Gaussian distribution.
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