Probabilities and Conditional Expectation

Question

Good day! Please check my answers. Here is the problem: Let $ X,Y, Z$ have joint pdf $f(x,y,z) = \frac23 (x+y+z), 0<x<1, 0<y<1,0<z<1,$ zero elsewhere. Find the marginal probability density functions.$$ f_x(x) = \int\limits_0^1 \int\limits_0^1\frac23 (x+y+z) \ dy \ dz = \frac{2x+3} 3, 0<x<1$$ $$f_y(y) = \int\limits_0^1 \int\limits_0^1\frac23 (x+y+z) \ dx \ dz = \frac{2y+3} 3, 0<y<1$$ $$f_z(z) = \int\limits_0^1 \int\limits_0^1\frac23 (x+y+z) \ dx \ dy = \frac{2z+3} 3, 0<y<1 $$$P(0<X<\frac12,0<Y<\frac12,0<Z<\frac12)= \int\limits_0^\frac12\int\limits_0^\frac12 \int\limits_0^\frac12\frac23 (x+y+z) \ dx \ dy \ dz=\frac1{12}$How about this one. How will I answer this one? Compute for $P(0<X<\frac12)=P( 0<Y<\frac12)=P(0<Z<\frac12)$ For the conditional expectation of $E(X^2YZ+3XY^4Z^2)= E(X^2YZ)+3E(XY^4Z^2)= \int\limits_0^1\int\limits_0^1\int\limits_0^1 (x^2yz)\frac23 (x+y+z)+ 3\int\limits_0^1\int\limits_0^1\int\limits_0^1 (xy^4z^2)\frac23 (x+y+z) \ dx \ dy \ dz= \frac{574}{2160} $

Answer 1

Unfortunately, your solution is not correct. The integrals are $f_{x}(x)=\int^{1}_{0} \int^{1}_{0} \frac{2}{3}(x + y +z) dydz = \frac{2x+2}{3}, x \in (0;1)$

and symetrically the second and the third double-integrals. What you got are even not densities because integrals of them is not 1 in any case. You can compute these double-integral iteratelly by using Fubini's Theorem.

Your idea how to compute $P(X < \frac{1}{2}, Y < \frac{1}{2}, Z < \frac{1}{2})$ is good but you did not compute the triple-integral correctly. It is

$\int^{\frac{1}{2}}_{0} \int^{\frac{1}{2}}_{0} \int^{\frac{1}{2}}_{0} \frac{2}{3}(x + y + z) dxdydz = \frac{1}{16}$.

You can also compute it by using Fubini's Theorem.