Independence of Random Variables: Discrete and Continuous Cases - Prof. Galen Shorack, Study notes of Probability and Statistics

The concept of independent random variables in both discrete and continuous cases. The necessary and sufficient conditions for independence, provides examples of buffon's needle and poisson distributions, and discusses computing marginal densities and conditional probabilities. The document also touches upon the concept of location/scale families.

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Lecture 10: Jan 28. Independent random variables Ross 6.2
Xand Yare independent if for any subsets Aand Bof ,P(XAYB) = P(XA)×P((YB).
10.1 Independence of two r.vs: discrete case
Discrete case: this is equivalent to pX,Y (x, y) = P(X=x, Y =y) = P(X=x).P (Y=y) = pX(x)pY(y).
Clearly this is necessary: take A={x}and B={y}.
Conversely, if pX,Y (x, y) = pX(x)pY(y), then for any A,B:
P(XA, Y B) = X
xAX
yB
pX,Y (x, y) = X
xAX
yB
pX(x)pY(y)
=X
xA
pX(x)X
yB
pY(y) = P(XA)P(YB)
Note this must hold for all x, y. Thus the ranges of the r.v.s cannot depend on each other. Example: Xis
value on first die, Yon second: these are independent. But if we throw out doubles (i.e. points with x=y)
they are no longer independent: if X= 4 we know Y6= 4.
10.2 Independence of two r.vs: continuous case
With A= (−∞, x) and B= (−∞, y) we see FX,Y (x, y) = FX(x)FY(y).
As with the 1-dimensional case, this is also sufficient:
Xand Yare independent if and only if FX,Y (x, y) = FX(x)FY(y) for all x, y .
Differentiating, we see this means fX,Y (x, y) = fX(x)fY(y), and conversely integrating we see these are
equivalent. Note again it must hold for all x, y: the ranges of the r.vs cannot depend on each other.
10.3 Buffon’s needle: a classic example of estimating π.
Parallel lines distance Dapart; needle length L, with LD.
Let Xbe distance from needle midpoint to nearest line; θbe angle of needle to X.Xis U(0, D/2), θis
U(0, π/2) and they are independent. We want the probability
P(X(L/2) cos(θ)) = Z Z2x<L cos(y)
fX(x)fθ(y)dx dy =2
π
2
DZπ/2
0ZL/2 cos(y)
0
dx dy
=4
π D Zπ/2
0
L
2cos(y)dy =2L
π D[sin(y)]π /2
0= 2L/πD
10.4 Examples of independence and dependence
Xand Yare independent if and only if fX,Y =fX(x)fY(y) for all x, y.
Also, if fX,Y =g1(x)g2(y) for all X, Y , then Xand Yare independent, and fX(x)g1(x), fY(y)g2(y).
(i) Example: fX,Y (x, y ) = x+yon 0 < x < 1, 0 < y < 1. (Ch 6: # 22)
Xand Yare not independent: why not?
(ii) Example: fX,Y (x, y ) = exp((x+y)) on 0 < x < , 0 < y < .
Xand Yare independent: why? What is fX(x) ?
(iii) Example: fX,Y (x, y ) = 2 exp((4x+1
2y)) on 0 < x < , 0 < y < .
Xand Yare independent: why? What is fX(x) ?
(iv) Example: fX,Y (x, y ) = 1/6 on 0 < x < 2, 0 < y < 3,
Xand Yare independent: why? What is fX(x) ?
(v) Example: fX,Y (x, y ) = 2 on 0 < x, 0 < y,x+y < 1
Xand Yare not independent: why not?
1
pf3

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Lecture 10: Jan 28. Independent random variables Ross 6.

X and Y are independent if for any subsets A and B of ℜ, P (X ∈ A ∩ Y ∈ B) = P (X ∈ A) × P ((Y ∈ B).

10.1 Independence of two r.vs: discrete case Discrete case: this is equivalent to pX,Y (x, y) = P (X = x, Y = y) = P (X = x).P (Y = y) = pX (x)pY (y).

Clearly this is necessary: take A = {x} and B = {y}. Conversely, if pX,Y (x, y) = pX (x)pY (y), then for any A, B:

P (X ∈ A, Y ∈ B) =

∑ x∈A

∑ y∈B

pX,Y (x, y) =

∑ x∈A

∑ y∈B

pX (x)pY (y)

=

∑ x∈A

pX (x)

∑ y∈B

pY (y) = P (X ∈ A) P (Y ∈ B)

Note this must hold for all x, y. Thus the ranges of the r.v.s cannot depend on each other. Example: X is value on first die, Y on second: these are independent. But if we throw out doubles (i.e. points with x = y) they are no longer independent: if X = 4 we know Y 6 = 4.

10.2 Independence of two r.vs: continuous case With A = (−∞, x) and B = (−∞, y) we see FX,Y (x, y) = FX (x)FY (y). As with the 1-dimensional case, this is also sufficient: X and Y are independent if and only if FX,Y (x, y) = FX (x)FY (y) for all x, y.

Differentiating, we see this means fX,Y (x, y) = fX (x)fY (y), and conversely integrating we see these are equivalent. Note again it must hold for all x, y: the ranges of the r.vs cannot depend on each other. 10.3 Buffon’s needle: a classic example of estimating π.

Parallel lines distance D apart; needle length L, with L ≤ D. Let X be distance from needle midpoint to nearest line; θ be angle of needle to X. X is U (0, D/2), θ is U (0, π/2) and they are independent. We want the probability

P (X ≤ (L/2) cos(θ)) =

∫ ∫ 2 x<L cos(y)

fX (x)fθ(y) dx dy = (^) π^2 D^2

∫ (^) π/ 2 0

∫ (^) L/2 cos(y) 0

dx dy

= 4 π D

∫ (^) π/ 2 0

L

cos(y) dy = 2 L π D [sin(y)]π/ 0 2 = 2 L/πD

10.4 Examples of independence and dependence

X and Y are independent if and only if fX,Y = fX (x)fY (y) for all x, y. Also, if fX,Y = g 1 (x)g 2 (y) for all X, Y , then X and Y are independent, and fX (x) ∝ g 1 (x), fY (y) ∝ g 2 (y).

(i) Example: fX,Y (x, y) = x + y on 0 < x < 1, 0 < y < 1. (Ch 6: # 22) X and Y are not independent: why not?

(ii) Example: fX,Y (x, y) = exp(−(x + y)) on 0 < x < ∞, 0 < y < ∞. X and Y are independent: why? What is fX (x)?

(iii) Example: fX,Y (x, y) = 2 exp(−(4x + 12 y)) on 0 < x < ∞, 0 < y < ∞. X and Y are independent: why? What is fX (x)?

(iv) Example: fX,Y (x, y) = 1/6 on 0 < x < 2, 0 < y < 3, X and Y are independent: why? What is fX (x)?

(v) Example: fX,Y (x, y) = 2 on 0 < x, 0 < y, x + y < 1 X and Y are not independent: why not?

Lecture 11: Jan 30. Examples of computing probabilities

11.1 Computing marginal densities

(i) Example: f (x, y) = x + y on 0 < x < 1, 0 < y < 1. (Ch 6: # 22)

What is fX (x)?

(ii) Example: fX,Y (x, y) = 2 on 0 < x, 0 < y, x + y < 1 X and Y are not independent: why not?

What is fX (x)? (ii) Example: fX,Y (x, y) = 2 exp(−(4x + 12 y)) on 0 < x < ∞, 0 < y < ∞. X and Y are independent: why? What is fX (x)? How would you compute fX (x) if you did not see X and Y are independent?

11.2 Computing probabilities with joint pdf ’s

Recall: P ((X, Y ) ∈ A) = ∫ ∫ A fX,Y^ (x, y)^ dx dy. (i) Example: f (x, y) = x + y on 0 < x < 1, 0 < y < 1. (Ch 6: # 22)

P (X + Y < 1) =

∫ (^1) y=

∫ (^1) −y x=

(x + y) dx dy =

∫ (^1) y=

[^1

x^2 + yx]^10 − ydy

=

∫ (^1) y=

(^1

(1 − y)^2 + y(1 − y))dy =

∫ (^1) y=

(1 − y^2 )dy = 1 2

(ii) Example: f (x, y) = 2 on 0 < x, 0 < y, x + y < 1

P (X < 3 Y ) =

∫ (^1) y=

∫ (^) min(1−y, 3 y) x=

2 dx dy =

∫ (^1) / 4 y=

6 y dy +

∫ (^1) y=1/ 4

2(1 − y) dy

= [3y^2 ]^10 / 4 + [−(1 − y)^2 ]^11 / 4 = 3 /16 + 9/ 16 = 3 / 4.

11.3: Sum of two independent U(0,1)

Let X ∼ U (0, 1) and Y ∼ U (0, 1) with X and Y independent. So fX,Y (x, y) = 1 on 0 < x < 1, 0 < y < 1.

(a) What is the range of W = X + Y?

(b) Compute P (X + Y < w) where 0 < w < 1. Or, draw a picture. (Answer: w^2 /2)

(c) Hence show fW (w) = w if 0 < w < 1. (This is only part of fW (w).) (c) Note (2 − W ) = (1 − X) + (1 − Y ). Why does this tell me that fW (w) is symmetric about w = 1? (d) Because of its shape, fW (w) is known as a triangular density. Do these densities form a location/scale family?