SOLUTION FOR HOMEWORK 11, ACTS 4306

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1 SOLUTION FOR HOMEWORK, ACTS 36 Welcome to your th homework. This is a collection of transformation, Central Limit Theorem (CLT), and other topics.. Solution: By definition of Z, Var(Z) = Var(3X Y.5). We know that Var(aU + b) = a 2 Var(U) for any constants a and b. Further, for independent U and V we have Var(U + V ) = Var(U) + Var(V ). Using this we continue Var(Z) = Var(3X Y ) = 9Var(X) + ( ) 2 Var(Y ) = (9)() + 2 =. Answer: D 2. Solution: This is one-variable transformation problem. Function g(t) = t 2 = y is one-to-one for t > 2. As a result, the inverse function g (y) = y /2 exists for y >. Then we use Theorem 7., f Y (y) = dg (y)/dy f T (g (y)). Now we note that and this yields f T (t) = df T (t)/dt = (8/t 3 )I(t > 2), f Y (y) = dy /2 /dy [(8/t 3 )I(t > 2)] t=y /2 = (/2)y /2 8y 3/2 I(y /2 > 2) = y 2 I(y > ). [Now please check that this density is integrated to!] 3. Notation: X and Y are profits of Companies I and II, respectively. Given: Y = 2X, f X (x). Find: f Y (y). Solution: Here Y = g(x) = 2X which is one-to-one transformation. Then f Y (y) = dg (y)/dy f X (g (y)) = d(y/2)/dy f X (y/2) = (/2)f X (y/2). Remark: Please remember that if Y = ax + b, which is a scale-location transformation, then f Y (y) = a f X ((y b)/a). If you know this fact then you solve this problem instantly.. Given: Y = X.8 and f X (x) = e x I(x > ). Find: f Y (y), y >. Solution: y = g(x) = x.8 is one-to-one function for x >. Further, g (y) = [y/] /.8 = [y/].25, and f Y (y) = dg (y)/dy f X ([y/].25 ) = (.25/)[y/].25 e (y/).25 I(y > ).

2 Answer: E 5. Solution: We need to find density of X = 2T +T 2. Let us use the convolution formula f U+V (z) = f U (u)f V (z u)du. (if you forgot it obtain it using one of our methods!) In the considered case U = 2T and then f 2T (u) = (/2)e u/2 I(u > ) and V = T 2 with f T2 (v) = e v I(v > ). Using these results we get for x > f 2T +T 2 (x) = (/2)e u/2 I(u > )e (x u) I(x u > )du x = e x [ (/2)e u/2 du] = e x [e u/2 ] x u= = e x [e x/2 ] = e x/2 e x. 6. Notation: T and T 2 are times until failure of the first and second generators, respectively. Solution: Because T and T 2 are independent, Var(T + T 2 ) = Var(T ) + Var(T 2 ) = = 2. Answer: E 7. Solution: Let X = Y/Z where Y and Z are supported on (, ). Introduce V := Y and consider the system { X = Y/Z V = Y which is equivalent to { Y = V Z = V/X The absolute value of the Jacobian for the last system is x v/x 2 = v/x2. Then we get f XV (x, v) = vx 2 f Y (v)f Z (v/x). In our particular case f Y (y) = e y I(y > ), f Z (z) = (/2)e z/2 I(z > ), and then using the above-obtained formula we get f XV (x, v) = vx 2 e v (/2)e (v/x)/2 I(x > )I(v > ). 2

3 Then f X (x) = (/2)x 2 ve v[+/(2x)] dv (remember that for an exponential random variable we know that its mean is x(/λ)e x/λ dx = λ) = 2x 2 [ + /2x] = 2 2 (2x + ) Given: Y = 225 l= X l where E(X) = 325 and Var(X) = (25) 2. Find: a constant y.9 such that P(Y y.9 ) =.9. Solution: Because the number of addends n = 225 > 3, we can use normal approximation meaning that we can assume that Y is Normal(µ := 325n, 2 := (25) 2 n). As a result using z-scoring and notation Z for standard normal random variable, we get P(Y y.9 ) = P( Y µ y.9 µ ) = P(Z y.9 µ ) =.9. We know from the Normal Table that P(Z.28) =.9, so we get y.9 (325)(225) 25(225) /2 =.28 and solving this equation gives us y.9 = 6, 32, 525. Answer: C 9. Notation: T is the total number of claims. Given: T = 25 l= X l where X l are independent Poisson(2). Find: P(25 < T < 26). Solution: This is again the normal approximation problem since n = 25 > 3. We have E(X l ) = 2, Var(X l ) = 2, and then T is Normal(µ = 2n, 2 = 2n). Denote by Z a standard normal random variable, and using z-scoring we get 25 nµ P(25 < T < 26) = P( < T nµ (plug-in numbers, calculate the ratios, and then use the Normal Table) = P( < Z < 2) =.9772 (.83) =.885. < 26 nµ ). Given: P(N = ) = /2, P(N = ) = /3, P(N > ) = /6, P(S = N = ) =, f S (s N = ) = (/5)e s/5 I(s > ), f S (s N > ) = (/8)e s/8 I(s > ). Find: P( < S < 8). 3

4 Solution: Use the total probability theorem and write P( < S < 8) = P(( < S < 8) (N = ))+P(( < S < 8) (N = ))+P(( < S < 8) (N > )) Answer: C = + P(N = )P( < S < 8 N = ) + P(N > )P( < S < 8 N > ) = (/3) 8 (/5)e s/5 ds + (/6) 8 (/8)e 5/8 ds = (/3)[e /5 e 8/5 ] + (/6)[e /8 e 8/8 ] =.23. Solution: This is the normal approximation problem since n = > 3. Let U = X + Y denote the time a person watches movies and sporting events. Then and E(U) = E(X) + E(Y ) = 7 Var(U) = Var(U) + Var(V ) + 2Cov(X, Y ) =. Then the total time T = n l= U l, n = is approximately normally distributed with mean µ = 7n and variance 2 = n. Using z-scoring and notation Z for a standard normal random variable we get = P(Z < P(T < 7) = P( T µ In the last step I used the Normal Table. < 7 µ ) 7 (7)() [()()] /2 = P(Z < ) = Notation: x is premium for a policy, C - total claims for a policy. Given: f C (y) = 3 e y/3 I(y > ), x = E(C) +, n = Find: P( n l= C l > x). Solution: This is again the Normal approximation problem. Note that µ = E(C) =, 2 = Var(C) = 2, x = + =, and then using z-scoring and notation Z for a standard normal random variable we write, P( n l= nl= nµ nµ C l > ) = P( > ) = P(Z > ) =.587. [n 2 ] /2 [n 2 ] /2 3. Given: X, X 2, X 3 are independent and distributed as X with f X (x) = 3x I(x > ), and X is in thousands. Find: E(max(X, X 2, X 3 )).

5 Solution: Denote Y = max(x, X 2, X 3 ). We need to find the distribution of Y and then we calculate its expectation. For the cdf we can write F Y (y) = P(max(X, X 2, X 3 ) y) = P((X y) (X 2 y) (X 3 y)) = [F X (y)] 3. Then F X (y) = y 3x dx = ( y 3 ) and we can write f Y (y) = df Y (y)/dy = 3F 2 X(y)f X (y) = 3( y 3 )(3y )I(y > ). Now we can answer the question of the problem: E(Y ) = 9 y 3 ( y 3 ) 2 dy = 9 = 9[(/2) 2(/5) + (/8)] = 2.3 [y 3 2y 6 + y 9 ]dy Because the amount is in thousands, the answer is (2.3)() = 23.. Notation: X and Y are times until failure for the two components. Find: P(min(X, Y ) < ). Solution: When we are dealing with the smallest random variable the approach is as follows: P(min(X, Y ) < ) = P(min(X, Y ) ) = P((X ) (Y )). (you can also get this using De Morgan Law). Now we can use the joint density, = We conclude that P((X ) (Y )) = Solution: By its definition 3 [(x + y)/27]dxdy = (/27) 3 = (/27) ( + 2y)dy = 6/27. f XY (x, y)dxdy [x 2 /2 + xy] 3 x= dy P(min(X, Y ) < ) = 6/27 = /27. M WZ (t, t 2 ) = E(e t W+t 2 Z ). Then, using notation W = X + Y and Z = Y X we get Answer: E M WZ (t, t 2 ) = E(e t (X+Y )+t 2 (Y X) ) = E(e (t +t 2 )Y )E(e (t t 2 )X ) = M Y (t + t 2 )M X (t t 2 ) = e (t +t 2 ) 2 /2 e (t t 2 ) 2 /2 = e t2 +t2 2. 5