Important results from last lecture(1) Joint Distribution of 2 random variables(2) Marginal distribution(3) Conditional distribution of Y given X=x(4) Variance, Covariance, Coefficient of correlation between 2 random variables(5) Moment generating function of a joint distributionMath 370X - Lecture 8Joint, Marginal and Conditional Distributions - Part IISaumil PadhyaOctober 31, 2016Important results from last lecture(1) Joint Distribution of 2 random variablesThe pdf of joint distribution is defined asDiscrete0 ≤ f(x, y) ≤ 1PxPyf(x, y) = 1Continuousf(x, y) ≥ 0∞R−∞∞R−∞f(x, y) dy dx = 1If X and Y are independent, thenf(x, y) = fX(x) · fY(y) − (1)The cdf of joint distribution is defined asF (x, y) = P [(X ≤ x) ∩ (Y ≤ y)]DiscreteF (x, y) =xPs=−∞yPt=−∞f(s, t)ContinuousF (x, y) =xR−∞yR−∞f(s, t) dt ds∂2∂x∂yF (x, y) = f(x, y)1If X and Y are independent, thenF (x, y) = FX(x) · FY(y) − (2)Expectation of joint distribution is defined asDiscreteE[h(X, Y )] =PxPyh(x, y) · f(x, y)ContinuousE[h(X, Y )] =∞R−∞∞R−∞h(x, y) · f(x, y) dy dxIf X and Y are independent, thenE[g(X)g(Y )] = E[g(X)] · E[g(Y )] − (3)(2) Marginal distributionfX(x) =Xyf(x, y) =∞Z−∞f(x, y) dyfY(y) =Xxf(x, y) =∞Z−∞f(x, y) dx(3) Conditional distribution of Y given X=xfY |X(y|X = x) =f(x, y)fX(x)f(x, y) = fY |Xy|X = x · fX(x)E[Y |X = x] =Zy · fY |X(y|X = x)dyE[Yk|X = x] =Zyk· fY |X(y|X = x)dyV ar[Y |X = x] = E[Y2|X = x] − (E[Y |X = x])2∗ ∗∞Z−∞fY |X(y|X = x)dy = 1 ∗ ∗2If X and Y are independent, then from (1) and (2)fY |X(y|X = x) =f(x, y)fX(x)=fX(x) · fY(y)fX(x)= fY(y) − (4)(4) Variance, Covariance, Coefficient of correlation between 2 random variablesCov[X, Y ] = E(X − E[X]) · E(Y − E[Y ]) = E[XY ] − E[X] · E[Y ]V ar[aX + bY + c] = a2· V ar[X] + b2· V ar[Y ] + 2ab · Cov[X, Y ]p(X, Y ) = pX,Y=Cov[X, Y ]σx· σywhere − 1 ≤ p(X, Y ) ≤ 1If X and Y are independent, then from (3)E[XY ] = E[X] · E[Y ] − (5)Cov[X, Y ] = E[XY ] − E[X] · E[Y ] = E[X] · E[Y ] − E[X] · E[Y ] = 0 − (6)V ar[aX + bY + c] = a2· V ar[X] + b2· V ar[Y ] + 2ab · Cov[X, Y ] = a2· V ar[X] + b2· V ar[Y ] − (7)(5) Moment generating function of a joint distributionMX,Y(t1, t2) = E[et1X+t2Y]E[Xn· Ym] =∂n+m∂nt1∂mt2MX,Y(t1, t2)t1=t2=0If X and Y are independent, thenMX,Y(t1, t2) = MX(t1) · MY(t2) − (8)The proof goes as followsMX,Y(t1, t2) = E[et1X+t2Y] =∞Z−∞∞Z−∞et1X· et2Y· f(x, y)dydx =∞Z−∞∞Z−∞et1X· et2Y· fX(x) · fY(y)dydxMX,Y(t1, t2) =∞Z−∞et1XfX(x)dx ·∞Z−∞et2YfY(y)dy = MX(t1) ·
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