# Recitation 4. 24xy for 0 < x < 1, 0 < y < 1, x + y < 1 0 elsewhere

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1 Recitation. Exercise 3.5: If the joint probability density of X and Y is given by xy for < x <, < y <, x + y < fx, y) elsewhere find PX + Y < ). The blue shaded region is the set of points x, y) s for which fx, y) > the joint probability density function is zero everywhere else), and the probability that we need to find is the integral of fx, y) over the portion of the blue region staying below the line x + y. Carrying out this integration, we obtain / x xy dy dx / / / ) x x dx ) x + x x dx 3x + x 3 x ) dx Exercise 3.53: If the joint density of X and Y is given by for < x < y, < y < fx, y) y elsewhere find the probability that the sum of the values of X and Y will exceed /.

2 The blue shaded region is the set of points x, y) s for which fx, y) > the joint probability density function is zero everywhere else). Note that the probability that we need to find is PX + Y > /), and this is given by the integral of fx, y) over the portion of the blue region staying above the line x + y. Alternatively, we can compute the integral of fx, y) over the lower portion of the blue region and subtract the value of the integral from one. That is, we compute / y / / dx dy + y / / dy + / y y ) dx dy y ) ) dy + ln/) ln/)) ) ln. 3. Exercise 3.5: Find the joint probability density of the two random variables X and Y whose joint distribution is given by e x ) e y ) for x >, y > F x, y) elsewhere We have fx, y) F x, y) x y wherever the partial derivatives exist. Partial differentiation yields x y F x, y) xy e x +y ) for x > and y > and elsewhere. So we have xy e x +y ) for x >, y > fx, y) elsewhere. Exercise 3.: Find k if the joint the probability distribution of X, Y, and Z is given by fx, y, z) kxyz

3 for x, ; y,, 3; z,. Clearly, k must be non-negative. To find its value, we use the condition that the sum of fx, y, z) values for x, y,, 3 and z, is equal to one. That is, we must have fx, y, z). Note that the sum can be computed as 3 x y z Hence k /5. fx, y, z) kxyz 3kxy 8kx 5k. x y z x y z x y x 5. Exercise 3.3: With reference to Exercise, find a) PX, Y, Z ). b) PX, Y + Z ). a) PX, Y, Z ) f,, ) + f,, ) 3/5 /8. b) PX, Y + Z ) f,, ) + f, 3, ) /5 7/7.. Exercise 3.8: If the joint probability density of X, Y, and Z is given by x + 3y + z) for < x <, < y <, < z < fx, y, z) 3 elsewhere find a) PX, Y, Z ). b) PX <, Y <, Z < ). a) We have b) We compute P X, Y, Z ) P X <, Y <, Z < ) fx, y, z) dz dz dz. fx, y, z) dz dy dx x + 3y + z) dz dy dx 3 x + 3y + z) dz dy dx x + 3 y + 8 x + ) dy dx ) dx 3 + ) 8. 3

4 7. Exercise 7 a-b): With reference to Exercise see page 9); find a) the marginal distribution of X; b) the marginal distribution of Y. a) Let gx) be the marginal distribution of X. Then, for x,,, we compute g) g) g) b) Let hy) be the marginal distribution of Y. Then, for y,,, 3, we compute h) h) + + h) h3) 8. Exercise 7 a-c): Given the joint probability distribution find fx, y, z) xyz, for x,, 3; y,, 3; z, 8 a) the joint marginal distribution of X and Y ; b) the joint marginal distribution of X and Z; c) the marginal distribution of X. a) Let gx, y) be the joint marginal distribution of X and Y. Then, gx, y) fx, y, z) z xyz 8 xy, for x,, 3; y,, 3. 3 b) Let hx, z) be the joint marginal distribution of X and Z. Then, we compute it as hx, z) fx, y, z) y z xyz 8 xz, for x,, 3; z,. 8 c) Let lx) the marginal distribution of X. We can compute it as lx) y z fx, y, z) y y z xyz 8 y xy 3 x, for x,, 3,

5 Equivalently, or lx) gx, y) y y xy 3 x, for x,, 3. lx) hx, z) z z xz 8 x, for x,, Exercises 7 a & 75 a: If the joint density of X and Y is given by x + y) for < x <, < y < fx, y) elsewhere find a) the marginal density of X; b) the marginal density of Y. a) Let gx) be the marginal density of X. Clearly, for x /, ), For x, ), we have gx) fx, y) dy gx) fx, y) dy. x + y) dy x + ) x + ). Hence, x + ) for < x < gx) elsewhere b) Let hy) be the marginal density of Y. Likewise, for y /, ), For y, ), we have hy) hy) fx, y) dx fx, y) dx. x + y) dx + y). Hence, + y) for < y < hy) elsewhere. Exercise 77: With reference to exercise 53, find 5

6 a) the marginal density of X; b) the marginal density of Y. a) Let gx) be the marginal density of X. Clearly, for x /, ), For x, ), we have gx) gx) fx, y) dy fx, y) dy. Note for TA s: mention that ln x <, for x, ).) gx) x y dy ln y ln x. x ln x for < x < elsewhere b) Let hy) be the marginal density of Y. Likewise, for y /, ), For y, ), we have Hence, hy) hy) hy) fx, y) dx. fx, y) dx y for < y < elsewhere dx. y

7 Recitation. Let X and Y be two discrete random variables with the joint probability distribution Find a) the marginal distribution of X; b) the conditional distribution of Y given X. fx, y) x + y), for x,, 3; y,. a) Let f X x), for x,, 3, be the marginal distribution of X. Then we have f X x) fx, y) y y x + y) x + 3), for x,, 3. b) To find the conditional distribution of Y given X, we compute f Y X y ) f, y), for y,. f X ) Note that f X ) must be different than zero for the conditional distribution to make sense. In this case, we have f X ) 5. Carrying out the computation, we obtain f Y X y ) f, y) f X ) + y) 5 + y), for y,. 5. Let X and Y be two continuous random variables with the joint probability density xy for < x <, < y <, x + y < fx, y) elsewhere Find a) the marginal density of Y ; b) the conditional density of X given Y /. Note for TA s: draw the x-y axis and illustrate the region over which fx, y) > a) Let f Y x) be the marginal density of Y. Clearly, for y /, ), For y, ), we have f Y y) fx, y) dx f Y y) y fx, y) dx. y xy dx y x dx y y)

8 Hence, y y) for < y < f Y y) elsewhere Note that in terms of the indicator function for y, ),) y) for y /, ) we can rewrite the marginal density of Y as f Y y) y y),) y), for < y <. b) The function f X Y x ) fx, ) f y ), for < x < gives the conditional density of X given Y. Note that f Y ) must be different from zero for this definition to make sense. Here, f Y ) 3. For x /, ), fx, ), and therefore f X Y x ). For x, ), we have fx, ) x, and this gives fx ) x 3 8x. f X Y x ) 8x for < x < elsewhere Once again, in terms of the indicator function for x, ), )x) for x /, ) we can rewrite the marginal density of Y as f X Y x ) 8x, )x), for < x <. Reminder: For a set A, the indicator function A x) is defined as for x A A x) for x / A. 3. Exercise 7 c-d): With reference to Exercise see page 9); find c) the conditional distribution of X given Y ; d) the conditional distribution of Y given X. In Recitation 5, we solved parts a) and b) as follows:

9 a) Let f X x) be the marginal distribution of X. Then, for x,,, we compute f X ) f X ) f X ) b) Let f Y y) be the marginal distribution of Y. Then, for y,,, 3, we compute f Y ) f Y ) + + f Y ) f Y 3) Let us now solve parts c) and d). c) Let fx, y) be the joint probability distribution of X and Y as given in the table in Exercise ). To find the conditional distribution of X given Y, we compute f X Y x ) fx, ), for x,,. f Y ) Once again, note that f Y ) must be different than zero for the conditional distribution to make sense. In this case, it does make sense since f Y ) ). Carrying out the computations, we obtain f, ) f X Y ) f Y ) f, ) f X Y ) f Y ) f, ) f X Y ) f Y ) d) To find the conditional distribution of Y given X, we compute f Y X y ) f, y), for y,,, 3. f X ) Again, f X ) must be different than zero for the conditional distribution to make sense. In this case, it does make sense since f X ) 7 5 ). Carrying out the computations, we obtain f, ) f Y X ) f X ) 7 5 f, ) f Y X ) f X ) 7 5 f, ) f Y X ) f X ) f, 3) f Y X 3 ) f X )

10 . Exercise 7 d-e): Given the joint probability distribution find fx, y, z) xyz, for x,, 3; y,, 3; z, 8 d) the conditional distribution of Z given X and Y ; e) the joint conditional distribution of Y and Z given X 3. In Recitation 5, we solved parts a), b), c) as follows. a) Let f X,Y x, y) be the joint marginal distribution of X and Y. Then, f X,Y x, y) fx, y, z) z xyz 8 xy, for x,, 3; y,, 3. 3 b) Let f X,Z x, z) be the joint marginal distribution of X and Z. Then, we compute it as f X,Z x, z) fx, y, z) y z xyz 8 xz, for x,, 3; z,. 8 c) Let f X x) the marginal distribution of X. We can compute it as Equivalently, f X x) y z fx, y, z) y y z xyz 8 y xy 3 x, for x,, 3, or f X x) gx, y) y y xy 3 x, for x,, 3. f X x) hx, z) z Let us now solve parts d) and e). z xz 8 x, for x,, 3. d) To find the conditional distribution of Z given X and Y, we compute f Z X,Y z, ) f,, z), for z,. f X,Y, ) f X,Y, ) must be different than zero for the conditional distribution to make sense. It is easy to check that f X,Y, ) /8.) Using the explicit form of fx, y, z), we obtain f Z X,Y z, ) z 8 8 z, for z,. 3

11 e) To find the joint conditional distribution of Y and Z given X 3, we compute f Y,Z X y, z 3) f3, y, z), for y,, 3; z,. f X 3) f X 3) must be different than zero for the conditional distribution to make sense. It is easy to check that f X 3) /.) Using the explicit form of fx, y, z), we obtain f Y,Z X y, z 3) 3yz 8 yz, for y,, 3; z, Exercises 7 b & 75 b: If the joint density of X and Y is given by x + y) for < x <, < y < fx, y) elsewhere find 7 b) the conditional density of Y given X /; 75 b) the conditional density of X given Y. Note for TA s: Remind the students that we can rewrite the joint density of X and Y in terms of the indicator function for < x <, < y <,),) x, y) elsewhere as fx, y) x + y),),)x, y) for < x <, < y <. This is in line with the definition of A x) given earlier; we simply consider A as a subset of R and x as a point in R. In this question, our set is, ), ). Note that, in this case, we can write the indicator function as the product of,) x) and,) y). However, this decomposition as a product) may not hold for all subset of R. Consider the set S x, y) R : < x <, < y <, x + y < } given in question above. We can not write for x, y) S S x, y) elsewhere as the product of B x) and C y) for some subsets B and C of R. end of the note) In Recitation 5, we solved part 7 a & 75 a as follows: 7 a) Let f X x) be the marginal density of X. Clearly, for x /, ), f X x) fx, y) dy. 5

12 For x, ), we have Hence, f X x) fx, y) dy x + y) dy x + ) x + ). f X x) x + ),)x), for < x < in terms of the indicator function,) x) for x, ) for x /, ) Clearly we can also write the marginal density of X as x + ) for < x < f X x) elsewhere 75 a) Let f Y y) be the marginal density of Y. Likewise, for y /, ), For y, ), we have Hence, f Y y) f Y y) fx, y) dx fx, y) dx. x + y) dx + y). f Y y) + y),)y) for < y <. Now let us solve 7 b & 75 b. 7 b) The function f Y X y ) ) f, y f X ), for < y < gives the conditional density of Y given X / again note that f X /) 3/ ). For y /, ), f, y), and therefore f Y Xy ). For y, ), we have f, y) + y), and this gives f Y X y ) + y) 3 ) 3 + y + y). f Y X y ) + y),)y), for < x <.

13 75 b) The function f X Y x ) fx, ) f Y ), for < x < gives the conditional density of X given Y again note that f Y ) / ). For x /, ), fx, ), and therefore f X Y x ). For x, ), we have fx, ) x + ), and this gives f X Y x ) x + ) x + ). f X Y x ) x + ),)x), for < x <.. Let X and Y be two continuous random variables with the joint density for < x < y, < y < fx, y) y elsewhere Find a) the conditional density of Y given X /. b) the conditional density of X given Y /. The marginal densities of X and Y are computed in Recitation 5 see the solution of Exercise 77). We obtained these functions last recitation as follows: 7 a) Let f X x) be the marginal density of X. Clearly, for x /, ), For x, ), we have f X x) f X x) fx, y) dy fx, y) dy. x y dy ln y ln x. x f X x) lnx),) x), for < x <. 75 a) Let f Y y) be the marginal density of Y. Likewise, for y /, ), For y, ), we have Hence, f Y y) f Y y) fx, y) dx. fx, y) dx y dx. y f Y y),) x), for < x <. Now let us solve our exercise. 7

14 a) The function ) f Y X y ) f, y ), for < y < f X gives the conditional density of Y given X / again note that f X /) ln ), Note for TA s: remind the students that ln >. For y /, ), f, y), and therefore f Y Xy ). For y, ), we have f, y) y, and this gives f Y X y ) y ln y ln. f Y X y ) y ln,)y), for < y <. b) The function f X Y x ) fx, ) f Y ), for < x < gives the conditional density of X given Y / again note that f Y /) ). For x /, ), fx, ), and therefore f X Y x ). For x, ), we have fx, ), and this gives f X Y x ). f X Y x ), )x), for < x <. 7. Exercise 78: With reference to Example see page 9), find a) the conditional density of X given X 3 and X 3 ; b) the joint conditional density of X and X 3 given X. a) In Example, the joint marginal density of X and X 3 is computed as x + ) e x 3 for < x <, x 3 > f X,X 3 x, x 3 ) elsewhere Note for TA s: If you are not convinced that the students have a good understanding of marginals if you think you have enough time), you may go over the derivation of the joint marginal density of X and X 3 given in Example. 8

15 The conditional density of X given X 3 and X 3 is given by the function f X X,X 3 x 3, ) f 3, x, ) f X,X 3 3, ), for < x <. Again, note that f X,X 3 3, ) 5 e.) For x /, ), f 3, x, ), and therefore f X X,X 3 x 3, ). For x, ), we have f 3, x, ) 3 + x )e, and this gives ) f X X,X 3 x 3, 3 + x )e 5 ) e x ) + 3x. 5 ) ) f X X,X 3 x 3, + 3x for < x < 5 elsewhere b) In Example, the marginal density of X is computed as x + for < x < gx ) elsewhere Note for TA s: If you are not convinced that the students have a good understanding of marginals if you think you have enough time), you may go over the derivation of the joint marginal density of X given in Example. The joint conditional density of X and X 3 given X is given by the function ) f X,X 3 X x, x 3 f, x, x 3 ) f X ), for < x < and < x 3 <. Again, note that f X ).) For x /, ) or x 3, f, x, x 3 ), and therefore f X,X 3 X x, x 3 ). For x, ) and x 3 >, we have f, x, x 3 ) + x )e x 3, and this gives ) f X,X 3 X x, x 3 + x )e x 3 ) + x e x 3. ) ) f X,X 3 X x, x 3 + x e x 3 for < x < and x 3 > elsewhere 8. Let X and Y be two continuous random variables whose joint distribution function is given by e x ) e y ) for x >, y > F x, y) elsewhere Determine whether X and Y are independent. 9

16 Let F X x) be the distribution function of X, and let F Y y) be the distribution function of Y. Note that we have F X x) F x, ) for all < x <, and F Y y) F, y) for all < y <. That is, we have e x ) for x > F X x) elsewhere e y ) for y > F Y y) elsewhere It is easy to verify that we have F x, y) F X x)f Y y), for all < x < and < y <. Hence, X and Y are independent. 9. With reference to Exercise, determine whether X and Y are independent. We already computed the marginal distribution f X x) of X and the marginal distribution f Y y) of Y. see the solution of Exercise 7 above). For example with x and y 3, we have f X ) 7 5 whereas f, 3). Hence X and Y are not independent. and f Y 3). Let X and Y be two independent random variables with the marginal densities e x for x > f X x) elsewhere e y for y > f Y y) elsewhere Find a) the distribution function of Z X + Y ; b) the density of Z. Note that since X and Y are independent, their joint density is given by the product of marginal densities. That is, if we let fx, y) be their joint density, then we have fx, y) f X x)f Y y) for all < x < and < y <. Using the marginal densities given in the question, we can write the function fx, y) as e x+y) for x > and y > fx, y) elsewhere a) Let F Z z) be the distribution of Z. Clearly Z takes values on, ). Hence, for z, F Z z).

17 For z >, we compute Hence we have F Z z) PZ z) PX + Y z) F Z z) x x f X,Y x, y)dy dx e x+y) dy dx x e x e y dy dx e x e z x) ) dx e x e z ) dx e z ze z. e z ze z for z > elsewhere Note that the distribution function for z > can also be obtained using the convolution formula discussed in class. That is; F Z z) f X x)f Y z x)dx, where F Y y) denotes the distribution function of Y. For y, F Y y) y f Y u) du is obviously zero, and for y > we have F Y y) Hence we have F Y y) e z ), ) y). f Y u)du e u du e z. When we go back to the convolution formula for the distribution function), we obtain F Z z) and this gives the same result for z >. f X x)f Y z x)dx e x e z x) )dx b) Let f Z z) the density of Z. Note that we have f Z z) F Z z) wherever F Zz) is differentiable. For z >, For z <, F Zx). F Zz) e z e z + ze z ze z The assignment at the point z does not matter. We can set it to zero for convenience. Hence, we write ze z for z > f Z z) for z The density function of Z can also be obtained directly without finding the distribution first. For z >, the density can be found using the convolution formula for the density function) as f Z z) f X x)f Y z x)dx e x e z x) dx e z dx ze z. Since both X and Y take values on, ), the density function f Z z) for z can be immediately written as zero.

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