2. The joint density for X and Y is: f(x, y) = 3(x + y) ...

2. The joint density for X and Y is: $f(x, y) = \begin{cases} 3(x + y) & x > 0, y > 0, x + y < 1 \\ 0 & \text{otherwise} \end{cases}$ a. Find $P(X < Y)$ b. Find the marginal density of Y. c. Find $P(Y < \frac{1}{2})$ d. Find $E(Y)$ and $Var(Y)$ e. Find the conditional density of X given Y. (Don't forget bounds.) f. Find $P(X < \frac{1}{3} | Y = \frac{1}{2})$ g. Find $E(X | Y = \frac{1}{2})$ h. Find $Var(X | Y = \frac{1}{2})$ i. Are X and Y independent? j. Find $Cov(X, Y)$

Transcript

00:01 Hello student, in the given question in part 8, sketch the region of non -zero probability density, non -zero probability, probability, density, density, we consider the joint probability density function as f of x, y which is equals to c into x, y square, 0 is less than x is less than 1 and 0 is less than y is less than 1 otherwise, so the region of non -zero probability density correspond to the range between where f of x, y is greater than, f of x, y is greater than 0, in this case the limit to the range is 0 is less than x is less than 1 and 0 is less than y is less than 1, so the region can be visualized as squared in the xy plane with sides of length to find the value of c, we need to ensure that the probability over the entire region is 1, so double integration of f of x, y of dx into dy which is equals to 1, then double integration of c of x into y bracket square dx into dy which is equals to 1, now integrate limits from 0 to 1 for both x and y, we get integration from 0 to 1, then again integration from 0 to 1, c of x into y square into dx into dy which is equals to 1, then c is constant outside integration from 0 to 1, y square into again integration from 0 to 1, x into dx into dy which is equals to 1, so therefore we get c of integration of 0 to 1, y square into 0 .5 into dy which is equals to 1 and c of 0 .5 is outside of integration 0 to 1, y square 0, y square y square dy which is equals to 1, so we get c of 0 .5 into 1 by 3 which is equals to 1, so c upon 6 equals to 1, c upon 6, c upon 6 equals to 1, so c equals to 6, therefore we have c is equals to 6, therefore we have c is equals to 6 and the region of non -zero probability density square with the sides of length 1 in xy plane and the region, the region of non -zero probability, non -zero probability density, probability density is the square, square with sides, with sides of length, of length 1 in xy plane.

03:16 In part b, we have to find the probability of p of x plus y is less than 1, p of x plus y is equals to 1 and p of, and p of x plus y is greater than 1, so we need to integrate the joint probability density function over the respective region, so probability of x plus y is less than 1 which is equals to double integration of f of x into y dx into dy over the region of x plus y is less than 1, similarly probability of x plus y is equals to 1 which is equals to double integration of f of x comma y dx into dy over the region of x plus y equals to 1 and third is probability of x plus y is greater than 1 which is equals to double integration of f of x comma y dx into dy.

04:18 The specific calculation required, then in part c, we have to find the marginal density function of x and y, so we integrate the joint probability density function over the respective two times, so the marginal density function of x comma f of x of x can be obtained by integrating f of x into x which is equals to integration of f of x comma y into dy integrating from 0 to 1 which is equals to put the function as cxy square into dy integrating from 0 to 1, then we get c outside because c is constant, so which is equals to c of integration from 0 to 1 x into y square into dy which is equals to c of 1 by 3 into x into y cube evaluating from 0 to 1, so which is equals to c of 1 by 3 of x into 1 cube minus 0 cube, so which is equals to which is equals to c of 1 by 3 of x...

Question

Please give Ace some feedback