By Gill D.M.

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**Example text**

The given probabilities are obtained by noting the following equivalences of events. 4 a. 1 2 0 0 C(x + 2y)dxdy = 4C = 1, thus C = 14 . 1 1 (x 0 4 + 2y)dy = 14 (x + 1) 0 < x < 2 0 otherwise x y c. FXY (x, y) = P (X ≤ x, Y ≤ y) = −∞ −∞ f (v, u)dvdu. The way this integral is calculated depends on the values of x and y. For example, for 0 < x < 2 and 0 < y < 1, b. fX (x) = x y x FXY (x, y) = y f (u, v)dvdu = −∞ −∞ 0 0 1 x2 y y 2 x (u + 2v)dvdu = + . 4 8 4 But for 0 < x < 2 and 1 ≤ y, x y FXY (x, y) = x 1 f (u, v)dvdu = −∞ −∞ 0 0 1 x2 x (u + 2v)dvdu = + .

Y > 0 . We now notice that √ ∞ 2 −y 2 /2 y e 2π 2 dy = 0 ∞ y > 0. Using the transforma- ∞ /β Thus EY = β 1/γ Γ( γ1 + 1) and VarY = β 2/γ Γ b. fx (x) = γ −y γ /β γ−1 y , βe 2 since √12π −∞ y 2 e−y /2 = 1, the variance of a standard normal, and the integrand is symmetric. Use integration-by-parts to calculate the second moment ∞ y 3 e−y EY 2 = 2 ∞ /2 ye−y dy = 2 0 2 /2 dy = 2, 0 where we take u = y 2 , dv = ye−y c. The gamma(a, b) density is 2 /2 . Thus VarY = 2(1 − π/4). fX (x) = 1 xa−1 e−x/b .

The term Thus A0 = limr→0 Ar = exp( n1 log( r−1 r−1 d d r r rxi = xi log xi since rxi = dr xi = dr exp(r log xi ) = exp(r log xi ) log xi = xri log xi . b. (i) if log Ar is nondecreasing then for r ≤ r log Ar ≤ log Ar , then elog Ar ≤ elog Ar . Therefore Ar ≤ Ar . Thus Ar is nondecreasing in r. (ii) d dr log Ar = −1 r2 log( n1 n x=1 xri ) + 1 r n 1 n rxr−1 i i=1 n 1 n xr i=1 i = r 1 r2 n i=1 n xri log xi x=1 xri − log( n1 where we use the identity for rxr−1 showed in a). i (iii) n r i=1 xi log xi n r x=1 xi r = log(n) + n − log( r n = log(n) + i=1 n = log(n) + i=1 n = log(n) − i=1 1 xr ) n x=1 i n r i=1 xi log xi n r x=1 xi xri n i=1 xri n i=1 xri n i=1 xri n xri ) − log( x=1 n xri r log xi − n i=1 xri xri ) log( x=1 n xri xri (r log xi − log( log( xri )) x=1 n x=1 xri xri n ) = log(n) − ai log( i=1 1 ).