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Lecture 20

The document is a lecture on Probability and Statistics for MATH F113, covering topics such as joint distributions, expected values, covariance, and correlation. It includes mathematical propositions, proofs, and examples related to random variables and their relationships. Key concepts discussed include the expectation of sums and products of random variables, properties of covariance and correlation, and their implications in statistical analysis.

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0% found this document useful (0 votes)
18 views17 pages

Lecture 20

The document is a lecture on Probability and Statistics for MATH F113, covering topics such as joint distributions, expected values, covariance, and correlation. It includes mathematical propositions, proofs, and examples related to random variables and their relationships. Key concepts discussed include the expectation of sums and products of random variables, properties of covariance and correlation, and their implications in statistical analysis.

Uploaded by

darshanrajagoli2
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Probability & Statistics

MATH F113
Semester I 2022-2023

Dr. Yasmeen Akhtar


BITS Pilani K K Birla Goa Campus
(Figure and data source: Text book, Reference book)
Example

Let X and Y be two continuous random variables where the support of X is


[0, 1] and the support of Y is also [0, 1]. If the joint cdf of X and Y i.e., the cdf
of bivariate rv (X, Y) is FX,Y (x, y) = xy then find the joint pdf of X and Y.
Expected Values of functions of jointly distributed random variables

Proposition: Let X and Y be jointly distributed rv’s with pmf p(x, y) or


pdf f (x, y) according to whether the variables are discrete or continuous.
Then the expected value of a function h(X, Y), denoted by E[h(X, Y)]
or µh(X,Y) , is given by
(
  h(x, y)p(x, y), if X and Y are discrete
E[h(X, Y)] = R •x Ry •
• • h(x, y)f (x, y) dx dy if X and Y are continuous.
Proof. Let Z = h(X, Y), then pmf of Z is

pZ (z) = P(Z = z) = P(h(X, Y) = z) = ÂÂ p(x, y)


(x,y):h(x,y)=z

  h(x, y)p(x, y) =  Â h(x, y)p(x, y)


x y z (x,y):h(x,y)=z

=Â ÂÂ z p(x, y)
z (x,y):h(x,y)=z

= Âz ÂÂ p(x, y)
z (x,y):h(x,y)=z

= Âz P(h(X, Y) = z)
z
= Âz P(Z = z)
z
= E[Z] = E[h(X, Y)].
Expectation of Sums of Random Variables

Suppose E[X] and E[Y] both are finite. Then,

E[X + Y] = E[X] + E[Y]

Similarly, if a and b are numerical constants then


E(aX + bY) = aE(X) + bE(Y)

Proof. Let h(X, Y) = aX + bY. Then,

E[aX + bY] = Â Â(ax + by)p(x, y)


x y

= Â Â ax p(x, y) + Â Â by p(x, y)
x y x y

= a  x pX (x) + b  y pY (y)
x y

= aE[X] + bE[Y]
Expectation of Product of Independent Random Variables

Suppose X and Y are independent. Then,

E[XY] = E[X]E[Y]

Proof.
E[XY] = Â Â xy p(x, y)
x y

= Â Â xy pX (x) pY (y)
x y

= Â x(Â y pY (y)) pX (x)


x y

= E[Y] Â x pX (x)
x
E[XY] = E[X]E[Y]
Covariance

The covariance between two rv’s X and Y is


Cov(X, Y) = E[(X µX )(Y µY )]
(
  (x µX )(y µY )p(x, y) X, Y discrete
= R •x Ry •
• • (x µX )(y µY )f (x, y) dx dy X, Y continuous

Cov(X, Y) = E[XY Y µX XµY + µX µY ]


= E(XY) E[Y]µX E[X]µY + µX µY
= E(XY) µY µX µX µY + µX µY
Cov(X, Y)= E(XY) µX µY = E(XY) E(X)E(Y)
Covariance

• If X and Y are independent then Cov(X, Y) = 0.


• The covariance Cov(X, Y) is the expected product of deviations of the
two variables from their respective mean values.

• Cov(X, X) = E[(X µX )2 ]= V(X)

• Cov(X, Y) = Cov(Y, X)

• Cov(X + Y, Z) = Cov(X, Z) + Cov(Y, Z)

• Cov(cX, Y) = c Cov(X, Y)

• Cov(a + bX, Y) = b Cov(X, Y)


Variance of Sums of Random Variables

If X and Y are random variables and a and b are numerical constants


then
V(aX + bY) = a2 V(X) + 2abCov(X, Y) + b2 V(Y)
If X and Y are independent then V(aX + bY) = a2 V(X) + b2 V(Y)
Covariance to measure the directional relationship

Fig: p(x, y) = 1/10 for each of ten pairs corresponding to indicated points: (a) positive covariance; (b) negative
covariance; (c) covariance near zero
Example
A nut company markets cans of deluxe mixed nuts containing almonds,
cashews, and peanuts. The net weight of each can is exactly 1 lb where the
weight contribution of each type of nut is random. Let X = weight of
almonds in a selected can, and Y = weight of cashews in a selected can
1 What is the region of positive density?

Suppose the joint pdf for (X, Y) is


(
24xy, 0  x  1, 0  y  1, x + y  1
f (x, y) =
0 otherwise.

2 What is probability that almonds and cashews together make up at


most 50% of the can?
3 Are X and Y independent r.v’s? Find the Cov(X, Y).
Covariance Drawback

• The computed value of covariance depends critically on the units of


measurement. Ideally, the choice of units should have no effect on a
measure of strength of relationship.

Let X = Policy 1 deductible amount


Y = Policy 2 deductible amount

p(x, y) y p(x, y) y
0 100 200 0 1 2
x 100 .20 .10 .20 x 1 .20 .10 .20
250 .05 .15 .30 2.5 .05 .15 .30
Correlation

The correlation coefficient of X and Y, denoted by Corr(X, Y), rX,Y ,


or just r, is defined by

Cov(X, Y)
rX,Y =
sX sY

Properties:
1. For any two rv’s X and Y, 1  r  1:
⇣ ⌘
0  V sXX + sYY = V(X)
s2
+ V(Y)
s2
+ 2Cov(X,Y)
sX sY = 2(1 + r)
X Y

) 1r
⇣ ⌘
V(X)
0  V sXX Y
sY = sX2
+ V(Y)
s2
2Cov(X,Y)
sX sY = 2(1 r)
Y

)r 1
2. The two variables are said to be uncorrelated (linearly) when r = 0.
Correlation: Properties (cont.)

3. If a and c are either both positive or both negative, then

Corr(aX + b, cY + d) = Corr(X, Y).

4. If X and Y are independent, then r = 0, but r = 0 does not imply


independence.
Example: Let X and Y be discrete rv’s with joint pmf p(x, y) = 1/4 for
(x, y) = ( 4, 1), (4, 1), (2, 2), ( 2, 2) and it is 0 o.w.
Correlation: Properties (cont.)

5. r = 1 or 1 iff Y = aX + b for some numbers a and b with a 6= 0.


6. r is actually not a completely general measure of the strength of a
relationship.
7. r is a measure of the degree of linear relationship between X and Y.
8. A value of r near 1 does not necessarily imply that increasing the
value of X causes Y to increase. It implies only that large X values are
associated with large Y values.
Correlation is NOT Causation!

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