virtual university of pakistan lecture no. 26 statistics and probability miss saleha naghmi...
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TOPICS FOR TODAY BIVARIATE Probability Distributions (Discrete and Continuous) Properties of Expected Values in the case of Bivariate Probability DistributionsTRANSCRIPT
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Virtual University of Pakistan
Lecture No. 26
Statistics and Probability
Miss Saleha Naghmi Habibullah
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IN THE LAST LECTURE, YOU LEARNT
• Mathematical Expectation, Variance & Moments of Continuous Probability Distributions
•Bivariate Probability Distribution (Discrete case)+
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TOPICS FOR TODAY•BIVARIATE Probability Distributions (Discrete and Continuous)
• Properties of Expected Values in the case of Bivariate Probability Distributions
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You will recall that, in the last lecture we began the discussion of the example in which we were drawing 2 balls out of an urn containing 3 black, 2 red and 3 green balls, and you will remember that, in this example, we were interested in computing quite a few quantities.
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I encouraged you to compute the probabilities of the various possible combinations of x and y values, and, in particular, I motivated you to think about the 3 probabilities that were equal to zero.
Let us re-commence the discussion of this particular example:
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EXAMPLE:
An urn contains 3 black, 2 red and 3 green balls and 2 balls are selected at random from it. If X is the number of black balls and Y is the number of red balls selected, then find
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i) the joint probabilityfunction f(x, y);
ii) P(X + Y < 1); iii) the marginal p.d. g(x)
and h(y);iv) the conditional p.d. f(x | 1),v) P(X = 0 | Y = 1); andvi) Are x and Y independent?
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As indicated in the last lecture, using the rule of combinations in conjunction with the classical definition of probability, the probability of the first cell came out to be 3/28.
By similar calculations, we obtain all the remaining probabilities, and, as such, we obtain the following bivariate table:
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Joint Probability DistributionY
X0 1 2 P(X = xi)
g(x)
0 3/28 6/28 1/28 10/281 9/28 6/28 0 15/282 3/28 0 0 3/28
P(Y = yj)h(y) 15/28 12/28 1/28 1
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This joint p.d. of the two r.v.’s (X, Y) can be represented by the formula
2,1,0x
.2yx028 2,1,0yy,xf
3yx2
2y
3x
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ii) To compute P(X + Y < 1), we see that x + y < 1 for the cells (0, 0), (0, 1) and (1, 0).
Therefore
P(X + Y < 1) = f(0, 0) + f(0, 1) + f(1, 0) = 3/28 +6/28 + 9/28 = 18/28 = 9/14
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3 2 3(0,0)0 0 2
f
82
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iii) The marginal p.d.’s are:
x 0 1 2g(x) 10/28 15/28 3/28
y 0 1 2h(x) 15/28 12/28 1/28
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iv) By definition, the conditional p.d. f(x | 1) is
f(x | 1) = P(X = x | Y = 1)
1h
1,xf1YP
1YandxXP
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N o w
73
2812
028
628
6
1,xf1h2
0x
T h e r e f o r e
2,1,0x,1,xf73
1h1,xf1|xf
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T h a t i s ,
21
286
371,0f
371|0f
21
286
371,1f
371|1f
00371,2f
371|2f
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Hence the conditional p.d. of X given that Y = 1, is
x 0 1 2
f(x|1) 1/2 1/2 0
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v) Finally,
P(X = 0 | Y = 1)
= f(0 | 1) = 1/2
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v i ) W e f i n d t h a t f ( 0 , 1 ) = 6 / 2 8 ,
2810
281
286
283
y,0f0g2
0y
28120
286
286
1,xf1h2
0x
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N o w ,2812
2810
286
i . e . ,1h0g1,0f a n d t h e r e f o r e X a n d Y a r eN O T s t a t i s t i c a l l y i n d e p e n d e n t .
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Next, we consider the concept of BIVARIATE CONTINUOUS PROBABILITY DISTRIBUTION:
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CONTINUOUS BIVARIATE DISTRIBUTIONS:
The bivariate probability density function of continuous r.v.’s X and Y is an integrable function f(x,y) satisfying the following properties:
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i ) f ( x , y ) > 0 f o r a l l ( x , y )
i i ) and,1dydxy,xf
i i i )
.dxdyy,xf
dYc,bXaPd
c
b
a
i ) f ( x , y ) > 0 f o r a l l ( x , y )
i i ) and,1dydxy,xf
i i i )
.dxdyy,xf
dYc,bXaPd
c
b
a
i ) f ( x , y ) > 0 f o r a l l ( x , y )
i i ) and,1dydxy,xf
i i i )
.dxdyy,xf
dYc,bXaPd
c
b
a
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Let us try to understand the graphic picture of a bivariate continuous probability distribution:
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The region of the XY-plane depicted by the interval
(x1 < X < x2; y1 < Y < y2) is shown graphically:
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(x1, y2)
(x1, y1)
y2
y1
0 x1 x2X
Y
(x2, y2)
(x2, y1)
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Just as in the case of a continuous univariate situation, the probability function f(x) gives us a curve under which we compute areas in order to find various probabilities, in the case of a continuous bivariate situation, the probability function f(x,y) gives a SURFACE
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and, when we compute the probability that our random variable X lies between x1 and x2 AND, simultaneously, the random variable Y lies between y1 and y2, we will be computing the VOLUME under the surface given by our probability function f(x, y) encompassed by this region.
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The MARGINAL p.d.f. of the continuous r.v. X is
dyy,xfxg
and that of the r.v. Y is
dxyxfyh ,
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That is, the marginal p.d.f. of any of the variables is obtained by integrating out the other variable from the joint p.d.f. between the limits – and +.
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The CONDITIONAL p.d.f. of the continuous r.v. X given that Y takes the value y, is defined to be
,yhy,xfy|xf
where f(x,y) and h(y) are respectively the joint p.d.f. of X and Y, and the marginal p.d.f. of Y, and h(y) > 0.
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Similarly, the conditional p.d.f. of the continuous r.v. Y given that X = x, is
provided that g(x) > 0.
,xgy,xfx|yf
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It is worth noting that the conditional p.d.f’s satisfy all the requirements for the UNIVARIATE density function.
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Finally:
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Two continuous r.v.’s X and Y are said to be Statistically Independent, if and only if their joint density f(x,y) can be factorized in the form f(x,y) = g(x)h(y) for all possible values of X and Y.
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EXAMPLE
Given the following joint p.d.f.
elsewhere,0
4, y 2 2; x 0 y),– – x (681 y)f(x,
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a) Verify that f(x,y) is a jointdensity function.
b) Calculate ,25Y,
23XP
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c) Find the marginal p.d.f. g(x) and h(y).
d) Find the conditional p.d.f.f(x | y) and f(y | x).
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SOLUTION
a) The joint density f(x,y) will be a p.d.f. if
(i) f(x,y) > 0 and
(ii) 1dydx)y,x(f
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Now f(x,y) is clearly greater than zero for all x and y in the given region, and
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dxdyyx681dydx)y,x(f
4
2
2
0
dx2
yxyy681
4
2
22
0
2
0
22
0xx6
81dxx26
81
141281
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Thus f(x,y) has the properties of a joint p.d.f.
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b) To determine the probability of a valueof the r.v. (X, Y)falling in the region X < 3/2 , Y < 5/2, we find
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dxdyyx681
25Y,
23XP
25
23
2y0x
= dx2
yxyy681 2
523
2
2
0
= 32
9x2x1564
1dx2x
815
81 2
323
02
0
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c) The marginal p.d.f. of
X is
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,dyy,xfxg
x
,dyyx681 4
2 2x0
4
2
2
2yxyy6
81
2x0
x341
, 2x0
= 0 , 2xOR0x
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S i m i l a r l y , t h e m a r g i n a l p . d . f . o f Y i s
dxyx681yh
2
0 , 4y2
y541
, 4y2
= 0 , e l s e w h e r e .
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d ) T h e c o n d i t i o n a l p . d . f . o f X g i v e nY = y , i s
yh
y,xfy|xf , w h e r e h ( y ) > 0
H e n c e
f ( x | y ) =
2x0,y52
yx6
y541
yx681
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a n d t h e c o n d i t i o n a l p . d . f . o f Y g i v e nX = x , i s
xg
y,xfx|yf , w h e r e g ( x ) > 0
H e n c e
x|yf
4y2,x32
yx6
x341
yx681
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Next, we consider two important properties of mathematical expectation which are valid in the case of BIVARIATE probability distributions:
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PROPERTY NO. 1
The expected value of the sum of any two random variables is equal to the sum of their expected values, i.e.
E(X + Y) = E(X) + E(Y).
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PROPERTY NO. 2
The expected value of the product of two independent r.v.’s is equal to the product of their expected values, i.e.
E(XY) = E(X) E(Y).
The result also holds for the difference of r.v.’s i.e.
E(X – Y) = E(X) – E(Y).
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It should be noted that these properties are valid for continuous random variable’s in which case the summations are replaced by integrals.
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Let us now verify these properties for the following example:
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EXAMPLE:
Let X and Y be two discrete r.v.’s with the following joint p.d.
x
y2 4
1 0.10 0.153 0.20 0.305 0.10 0.15
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Find E(X),
E(Y),
E(X + Y), and
E(XY).
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SOLUTIONTo determine the expected
values of X and Y, we first find the marginal p.d. g(x) and h(y) by adding over the columns and rows of the two-way table as below:
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x
y2 4 h(y)
1 0.10 0.15 0.253 0.20 0.30 0.505 0.10 0.15 0.25
g(x) 0.40 0.60 1.00
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Now E(X) = xj g(xj) = 2 × 0.40 + 4 × 0.60 = 0.80 + 2.40 = 3.2
E(Y) = yi h(yi) = 1 × 0.25 + 3 × 0.50 + 5 × 0.25= 0.25 + 1.50 + 1.25 = 3.0
Hence E(X) + E(Y) = 3.2 + 3.0 = 6.2
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i j
jiji y,xfyxYXE
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i j
jiji y,xfyxXYE
In order to compute E(XY) directly, we apply the formula:
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In the next lecture, we will discuss these two formulae in detail, and, interestingly, we will find that not only E(X+Y) equals E(X) + E(Y), but also E(XY) equals E(X) E(Y) implying that the random variables X and Y are statistically independent.
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IN TODAY’S LECTURE, YOU LEARNT
•BIVARIATE Probability Distributions (Discrete and Continuous)
• Properties of Expected Values in the case of Bivariate Probability Distributions
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IN THE NEXT LECTURE, YOU WILL LEARN
• Properties of Expected Values in the case of Bivariate Probability Distributions (Detailed discussion)
•Covariance & Correlation
•Some Well-known Discrete Probability Distributions:
•Discrete Uniform Distribution