Transcription of SOLUTION FOR HOMEWORK 5, STAT 4351
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SOLUTION FOR HOMEWORK 5, STAT 4351
SOLUTION FOR HOMEWORK 5, stat 4351 . Welcome to your 5th HOMEWORK , which finishes our study of chapter 3. Here we are exploring basics of multivariate random variables (rv). Now let us look at your problems. 1. Problem (a). By definition, F ( , .9) = P (X , Y .9) = P (X = 0, Y = 0) + P (X = 1, Y = 0) = 1/12 + 1/6. 2. Problem We know that the joint probability mass function should be summable to 1, and this will allow us to find c. Write, c(x2 + y 2). X X X. 1= f (x, y) =. x,y x { 1,0,1,3} y { 1,2,3}. [3x2 + (1 + 4 + 9)] = c[(3 + 0 + 3 + 27) + (4)(14)] = c[33 + 56] = 89c. X. =c x { 1,0,1,3}. Answer: c = 1/89. 3. Problem Here it is very important to write the density mathematically and then draw the support of the density together with the area of integration. We have: f (x, y) = 24xyI(0 < x < 1)I(0 < y < 1)I(x + y < 1). Then Z Z 1/2 Z 1/2 x P (X + Y < 1/2) = f (x, y)dxdy = [ f (x, y)dy]dx x+y<1/2 0 0.
Remark: It can be a good habit to check you answer via verifying that the marginal density is integrated to 1. Let us do this: Z ∞ fX(x)dx = Z 1 0 (1/4)(4x+2)dx
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