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1. (a) (b) The random variables X 1 and X 2 are independent, and each has p.m.f.f(x) = (x + 2) / 6 if x = –1, 0, 1. Find E(X 1 + X 2 ). E(X 1 ) = E(X 2 ) = 1 2 3 (–1) — + (0) — + (1) — = 6 6 6 1 — 3 E(X 1 ) + E(X 2 ) = 1 1 2 — + — = — 3 3 3 Find the p.m.f. of Y = X 1 + X 2, and use this p.m.f. to find E(Y). The space of Y is{–2, –1, 0, 1, 2} P(Y = –2) = P(X 1 = – 1 X 2 = – 1) = P(X 1 = – 1) P(X 2 = – 1) =(1/6)(1/6) = 1/36

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P(Y = –1) = P({X 1 = – 1 X 2 = 0} {X 1 = 0 X 2 = – 1}) = P(X 1 = – 1 X 2 = 0) + P(X 1 = 0 X 2 = – 1) = P(X 1 = – 1) P(X 2 = 0) + P(X 1 = 0) P(X 2 = – 1) = (1/6)(2/6) + (2/6)(1/6) = 4/36 = 1/9 P(Y = 0) = P({X 1 = – 1 X 2 = 1} {X 1 = 1 X 2 = – 1} {X 1 = 0 X 2 = 0}) = P(X 1 = – 1 X 2 = 1) + P(X 1 = 1 X 2 = – 1) + P(X 1 = 0 X 2 = 0) = P(X 1 = – 1) P(X 2 = 1) + P(X 1 = 1) P(X 2 = – 1) + P(X 1 = 0) P(X 2 = 0) = (1/6)(3/6) + (3/6)(1/6) + (2/6)(2/6) = 10/36 = 5/18 P(Y = 1) =P(Y = 2) =12/36 = 1/39/36 = 1/4

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The p.m.f. of Y is g(y) = 1/36if y = – 2 4/36 = 1/9if y = – 1 10/36 = 5/18if y = 0 12/36 = 1/3if y = 1 9/36 = 1/4if y = 2 E(Y) =2/3(as expected from part (a)) (c)What type of distribution does W = X 1 2 have? The space of W is{0, 1} P(W = 0) = P(W = 1) = P(X 1 = 0) =1/3 P(X 1 = – 1 X 1 = 1) = 2/3 W has aBernoulli distribution with p = 2/3.

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2.Suppose that the random variable X has a b(n 1, p) distribution, that the random variable Y has a b(n 2, p) distribution, and that the random variables X and Y are independent. What type of distribution does the random variable V = X + Y have? V has a b(n 1 + n 2, p) distribution. 3. (a) The random variables X 1 and X 2 are independent and respectively have p.d.f. f 1 (x) = 4x 1 3 if 0 < x 1 < 1andf 2 (x) = 2x 2 if 0 < x 2 < 1. Find the joint p.d.f. of (X 1, X 2 ). Since X 1 and X 2 are independent, their joint p.d.f. is f(x 1, x 2 ) = 8 x 1 3 x 2 if 0 < x 1 < 1, 0 < x 2 < 1

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Section 5.2 Suppose X 1 and X 2 are continuous type random variables with joint p.d.f. f(x 1, x 2 ) and with space S R 2. Suppose Y = u(X 1, X 2 ) where u is a function. A method for finding the p.d.f. of the random variable Y is the distribution function method. To use this method, one first attempts to find a formula for the distribution function of Y, that is, a formula for G(y) = Then, the p.d.f. of Y is obtained by taking the derivative of the distribution function, that is, the p.d.f. of Y is g(y) = P(Y y). G / (y). This method will generally involve working with a double integral. Suppose Y 1 = u 1 (X 1, X 2 ) and Y 2 = u 2 (X 1, X 2 ) where u 1 and u 2 are functions. A method for finding the joint p.d.f. of the random variables (Y 1, Y 2 ) is the change-of-variables method. This method can be used when the two functions define a one-to-one mapping between S R 2 and T R 2 as follows: y 1 = u 1 (x 1, x 2 ) y 2 = u 2 (x 1, x 2 ) x 1 = v 1 (y 1, y 2 ) x 2 = v 2 (y 1, y 2 )

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Then the space of (Y 1, Y 2 ) is T R 2, and the joint p.d.f. of (Y 1, Y 2 ) is g(y 1, y 2 ) = f[ v 1 (y 1, y 2 ), v 2 (y 1, y 2 ) ] | J | where J is defined to be the determinant of a Jacobian matrix as follows: J = det Each of the distribution function method and the change-of-variables method can be extended in a natural way to a situation where (X 1, X 2 ) is replaced with (X 1, X 2, …, X n ) for n > 2. v1v1——y1y2v2v2——y1y2v1v1——y1y2v2v2——y1y2—— Return to Class Exercise #3. J is called the Jacobian determinant.

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(b)Define the random variables Y 1 = X 1 2 and Y 2 = X 1 X 2. Use the change-of-variables method to find the joint p.d.f. of (Y 1, Y 2 ). First, we find the space of Y 1 = X 1 2 and Y 2 = X 1 X 2 as follows: x1x1 x2x2 (0,0) (0,1)(1,1) (1,0) y1y1 y2y2 (0,0) (1,1) (1,0) < y 1 <, < y 2 <010 y1y1 < y 2 <, < y 1 <01y22y22 1 or (Look at where the boundaries are mapped) y 1 = x 1 2 x 1 = y 2 = x 1 x 2 x 2 = y1y1 y 2 —— y 1 Then, we find the inverse transformation as follows:

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3. - continued Next, we find the Jacobian determinant as follows: The joint p.d.f. of Y 1 and Y 2 is J = det x1x1——y1y2x2x2——y1y2x1x1——y1y2x2x2——y1y2—— = 1 —– 2 y 1 det 0 – y 2 —– 2 y 1 3 1 —– y 1 = 1 — 2y 1 g(y 1, y 2 ) =8( ) 3 ( )= y1y1 y 2 —— y 1 1 — 2y 1 if 0 < y 1 < 1, 0 < y 2 < y 1 4y 2 ( or 0 < y 2 < 1, y 2 2 < y 1 < 1 )

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(c)Find the marginal p.d.f. for Y 1 and the marginal p.d.f. for Y 2. g 1 (y 1 ) = 0 y1y1 4y 2 dy 2 = y 2 = 0 y1y1 2y 2 2 = 2y 1 if 0 < y 1 < 1 g 2 (y 2 ) = y22y22 1 4y 2 dy 1 = y 1 = y 2 2 1 4y 2 y 1 =4y 2 – 4y 2 3 if 0 < y 2 < 1 To find g 2 (y 2 ), we first observe that the space 0 < y 1 < 1, 0 < y 2 < y 1 can be described as 0 < y 2 < 1, y 2 2 < y 1 < 1.

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4. (a) (b) The random variables X 1 and X 2 have joint p.d.f. f(x 1, x 2 ) = 8x 1 x 2 if 0 < x 1 < x 2 < 1. Are X 1 and X 2 independent random variables? Why or why not? Since the space of (X 1, X 2 ) is not rectangular, X 1 and X 2 cannot possibly be independent. Define the random variables Y 1 = X 1 / X 2 and Y 2 = X 2 – X 1. Use the change-of-variables method to find the joint p.d.f. of (Y 1, Y 2 ). First, we find the space of Y 1 = X 1 / X 2 and Y 2 = X 2 – X 1 as follows: x1x1 x2x2 (0,0) (0,1)(1,1) y1y1 y2y2 (0,0) (0,1) (1,0) < y 1 <, < y 2 <0101 – y 1 < y 2 <, < y 1 <0101 – y 2 (Look at where the boundaries are mapped) or

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x 1 y 1 = —x 1 = x 2 y 2 = x 2 – x 1 x 2 = y 1 y 2 —— 1 – y 1 y 2 —— 1 – y 1 Then, we find the inverse transformation as follows: Next, we find the Jacobian determinant as follows: J = det x1x1——y1y2x2x2——y1y2x1x1——y1y2x2x2——y1y2—— = y 2 ——— (1 – y 1 ) 2 det y 1 —— 1 – y 1 y 2 ——— (1 – y 1 ) 2 1 —— 1 – y 1 = y 2 ——— (1 – y 1 ) 2

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8( )( )= y 1 y 2 —— 1 – y 1 y 2 —— 1 – y 1 y 2 ——— (1 – y 1 ) 2 if 0 < y 1 < 1, 0 < y 2 < 1 – y 1 8 y 1 y 2 3 ——— (1 – y 1 ) 4 ( or 0 < y 2 < 1, 0 < y 1 < 1 – y 2 ) 4. - continued The joint p.d.f. of Y 1 and Y 2 isg(y 1, y 2 ) = (c)Find the marginal p.d.f. for Y 1 and the marginal p.d.f. for Y 2. g 1 (y 1 ) = 0 1 – y 1 8 y 1 y 2 3 ——— dy 2 = (1 – y 1 ) 4 y 2 = 0 1 – y 1 2 y 1 y 2 4 ——— = (1 – y 1 ) 4 2y 1 if 0 < y 1 < 1

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g 2 (y 2 ) = 0 1 – y 2 8 y 1 y 2 3 ——— dy 1 = (1 – y 1 ) 4 1 1 8 y 2 3 ——— –——— dy 1 = (1 – y 1 ) 4 (1 – y 1 ) 3 To find g 2 (y 2 ), we first observe that the space 0 < y 1 < 1, 0 < y 2 < 1 – y 1 can be described as 0 < y 2 < 1, 0 < y 1 < 1 – y 2. 0 1 – y 2 y 1 = 0 1 – y 2 8 – 12y 2 + 4y 2 3 ——————if 0 < y 2 < 1 3 3y 1 – 1 8 y 2 3 ———— = 6(1 – y 1 ) 3 Study Example 5.2-3 in the textbook, and compare this with the distribution function method for this same situation, suggested in Text Exercise 5.2-6. 5.

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The random variables X 1 and X 2 are independent and each has a gamma(1,2) (or 2 (2) or exponential(2)) distribution. 6. (a) (b) Find the joint p.d.f. of (X 1, X 2 ). Since X 1 and X 2 are independent, their joint p.d.f. is 1 x 1 + x 2 f(x 1, x 2 ) = — exp – ——— if 0 < x 1, 0 < x 2 4 2 Define the random variable Y = X 1 + X 2. Use the distribution function method to find the p.d.f. of Y. The space of Y = X 1 + X 2 is{y | 0 < y} The distribution function of Y = X 1 + X 2 is G(y) = P(Y y) = P(X 1 + X 2 y) =P(X 1 y – X 2 ) =P(0 < X 1 y – X 2, 0 < X 2 < y) =

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0 y 0 y – x 2 1 x 1 + x 2 — exp – ——— dx 1 dx 2 = 4 2 0 y 0 y – x 2 1 x 2 — exp – — dx 2 = 2 2 0 y 1 x 2 — exp – — dx 2 = 2 2 x 1 = 0 y – x 2 x 1 – exp – — 2 1 x 1 — exp – — dx 1 2 2 0 y 1 x 2 — exp – —dx 2 = 2 2 y – x 2 1 – exp – ——– 2

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0 y 1 x 2 — exp – — dx 2 = 2 2 y – exp – — 2 x 2 = 0 y x 2 – exp – — = 2 x 2 y – — exp – — 2 2 y 1 – exp – — 2 y – — exp – — 2 2 The p.d.f. of Y is g(y) = G / (y) = y — exp – — if 0 < y 4 2 We recognize that Y has a distribution. gamma(2,2) (or 2 (4))

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The random variables U and V are independent and have respectively a 2 (r 1 ) distribution and a 2 (r 2 ) distribution. 7. (a) (b) Find the joint p.d.f. of (U, V). Define the random variable W =. Use the distribution function method to find the p.d.f. of W by completing the steps outlined, after first reviewing the following two facts from calculus: (1) Since U and V are independent, their joint p.d.f. is u v u + v f(u,v) = ————————— exp – ——— if 0 < u, 0 < v (r 1 /2) (r 2 /2) 2 2 U / r 1 —— V / r 2 d — dy (r 1 +r 2 )/2 r 1 /2 – 1r 2 /2 – 1 a b g(x,y) dx = —y —y a b g(x,y) dx

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Example: d — dy –1 3 (x 3 y + xy 2 ) dx = —y —y (x 3 y + xy 2 ) dx –1 3 x 4 y x 2 y 2 — + —— 4 2 x = –1 3 d — dy –1 3 (x 3 + 2xy) dx (20y + 4y 2 ) d — dy 20 + 8y x 4 — + x 2 y 4 x = –1 3 20 + 8y

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(2) Example: 7. - continued From the chain rule and the Fundamental Theorem of Calculus, we have that d — dt a h(t)h(t) g(x) dx = d — dt g(h(t)) h(t). d — dt –1 t + 3 t x 3 dx = (t + 3 t) 3 (1 + 3/(2 t)) x 4 —= 4 x = –1 d — dt t + 3 t (t + 3 t) 4 1 ———— – — = 4 4 d — dt (t + 3 t) 3 (1 + 3/(2 t))

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The space of W = is{w | 0 < w} The distribution function of W = is G(w) = P(W w) = U / r 1 —— V / r 2 U / r 1 —— V / r 2 P U / r 1 —— w = V / r 2 P U = r 1 wV —— r 2 0 0 1 —————— (r 1 /2) (r 2 /2) 2 u eu e du r 1 wv —— r 2 (r 1 +r 2 )/2 r 1 /2 – 1 – u/2 v edv r 2 /2 – 1 – v/2

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The p.d.f. of W is g(w) = G / (w) = 7. - continued 0 1 —————— (r 1 /2) (r 2 /2) v e dv = r 2 /2 – 1 – v/2 2 (r 1 +r 2 )/2 r 1 /2 – 1 r 1 wv —— r 2 exp r 1 wv – —— 2r 2 r 1 v — r 2 (To simplify this p.d.f., we could either (1) make an appropriate change of variables in the integral, as is done in Example 5.2-4 of the textbook, or (2) do some algebra to make the formula under the integral a p.d.f. which we know must integrate to (one) 1, as we shall do here.) 0 —————— (r 1 /2) (r 2 /2) r 1 /2 r1— r2 r1— r2 2 (r 1 +r 2 )/2 v (r 1 +r 2 )/2 – 1 w r 1 /2 – 1 exp r 1 wv + r 2 v – ———— 2r 2 dv =

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0 —————— (r 1 /2) (r 2 /2) r 1 /2 r1— r2 r1— r2 2 (r 1 +r 2 )/2 v (r 1 +r 2 )/2 – 1 w r 1 /2 – 1 exp r 1 wv + r 2 v – ———— 2r 2 dv = 0 ——————————— (r 1 /2) (r 2 /2) r 1 /2 r1— r2 r1— r2 w r 1 /2 – 1 [(r 1 + r 2 )/2] 2 (r 1 +r 2 )/2 dv = v (r 1 +r 2 )/2 – 1 exp [(r 1 + r 2 )/2] v – —————— 2 / (1 + r 1 w/r 2 ) ————————————— (r 1 /2) (r 2 /2) r 1 /2 r1— r2 r1— r2 w r 1 /2 – 1 [(r 1 + r 2 )/2] (1 + r 1 w/r 2 ) (r 1 +r 2 )/2 0 [2 / (1 + r 1 w/r 2 )] (r 1 +r 2 )/2 dv v (r 1 +r 2 )/2 – 1 exp [(r 1 + r 2 )/2] v – —————— 2 / (1 + r 1 w/r 2 ) This is the p.d.f. for a random variable having a (r 1 + r 2 )/2 2 / (1 + r 1 w/r 2 )gamma(,) distribution.

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————————————— (r 1 /2) (r 2 /2) r 1 /2 r1— r2 r1— r2 w r 1 /2 – 1 [(r 1 + r 2 )/2] (1 + r 1 w/r 2 ) (r 1 +r 2 )/2 if 0 < w This is the p.d.f. for a random variable having a Fisher’s f distribution with r 1 numerator degrees of freedom and r 2 denominator degrees of freedom. This distribution is important in some future applications of the theory of statistics. Since the integral must be equal to, then we now have thatone (1) the p.d.f. of W is g(w) =

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Suppose the random variable F has an f distribution with r 1 numerator degrees of freedom and r 2 denominator degrees of freedom. 8. (a) (b) (c) (d) (e) (f) If r 1 = 5 and r 2 = 10, then P(F < 3.33) =0.95 If r 1 = 5 and r 2 = 10, then P(F > 5.64) =0.01 f 0.025 (4, 8) =5.05 f 0.975 (4, 8) =1/f 0.025 (8, 4) = f 0.025 (8, 4) =8.98 f 0.975 (8, 4) =1/f 0.025 (4, 8) = 1/8.98 = 0.111 1/5.05 = 0.198

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1 Continuous Distributions ch3. 2 A random variable X of the continuous type has a support or space S that is an interval(possibly unbounded) or a.

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