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Example text

Find the simultaneous frequency of the 2-dimensional random variable (X 1 , Y2 ). 5. Find the frequency of Y2 . 6. Check if the random variables X1 and Y2 are independent. 1) Concerning X1 , P {X1 > x1 } = P {T1 > x1 ∧ T2 > x1 } = P {T1 > x1 } · P {T2 > x2 } = e−2ax1 , thus P {X1 ≤ x1 } = 1 − e−2ax1 , x1 > 0, and X1 is exponentially distributed of the frequency fX1 = 2a e−2ax1 , 0, x1 > 0, x1 ≤ 0, and mean 1 . com 30 for x2 > 0, 2. Maximum and minimum of random variables Random variables III and for x2 ≤ 0.

Com 59 1 < Y2 < 3 3 2− 1 3 3 = + − 1 y2 1 3 fY2 (y2 ) dy2 3 = 2 1 1 5 − + 12 3 2 4. 1 Let (X, Y ) be a 2-dimensional random variable of frequency h(x, y) and marginal frequencies f (x) and g(y), and let f (x | y) be the conditional frequency of X, given Y = y. Let ϕ be a function : R → R, for which ∞ |ϕ(x)| f (x | y) dx < ∞ −∞ for alle y ∈ R. In such a case we deﬁne the conditional mean of ϕ(X), given Y = y, by ∞ (1) −∞ ϕ(x) f (x | y) dx. The conditional mean of ϕ(X), given Y , is the random variable, which for Y = y has the value of (1).

Conditional distributions Random variables III Alternatively we may use that E{X | Y } for Y = y has the value ∞ −∞ x f (x | y) dx, so E{E{X | Y }} = ∞ ∞ y=−∞ ∞ = x f (x | y) dx g(y) dy = −∞ ∞ x x=−∞ ∞ ∞ x x=−∞ y=−∞ f (x | y)g(y) dy dx ∞ f (x, y) dy dx = y=−∞ x f (x) dx = E{X}. −∞ 2) Then put ϕ(x) = (x − E{X | Y })2 . com 61 Click on the ad to read more 4. Conditional distributions Random variables III thus E{V {X | Y }} = ∞ −∞ ∞ 1 g(y) ∞ = −∞ −∞ ∞ x2 − 2x E{X | y} + (E{X | y})2 h(x, y) dx · g(y) dy −∞ x2 − 2x E{X | y} + (E{X | y})2 h(x, y) dx dy ∞ = E X2 − 2 −∞ ∞ ∞ −∞ x f (x | y) dx dy g(y) (E{X | y})2 dy + −∞ ∞ = E X2 − 2 −∞ ∞ + = E X2 − g(y) E{X | y} · −∞ ∞ −∞ g(y) · E{X | y} · E{X | y} dy g(y) (E{X | y})2 dy (E{X | y})2 g(y) dy = V {X} + (E{X})2 − ∞ (E{X | y})2 g(y) dy −∞ 2 = V {X} + (E{E{X | Y }}) − E{(E{X | Y })2 }, and hence V {X} = E{V {X | Y }} − (E{E{X | Y }})2 + E (E{X | Y })2 .

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