(a) Verify that f(x, y)={ 4 x y if 0 ⩽ x ⩽ 1,0 ⩽ y ⩽ 1 0 otherwise . is a joint d

(a) Verify that f(x, y)={ 4 x y if 0 ⩽ x ⩽ 1,0 ⩽ y ⩽...

(a) Verify that $$f(x, y)=\left\{\begin{array}{ll}{4 x y} & {\text { if } 0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1} \\ {0} & {\text { otherwise }}\end{array}\right.$$ is a joint density function. (b) If $X$ and $Y$ are random variables whose joint density func- tion is the function $f$ in part (a), find (i) $P\left(X \geqslant \frac{1}{2}\right) \quad$ (ii) $P\left(X \geqslant \frac{1}{2}, Y \leqslant \frac{1}{2}\right)$ (c) Find the expected values of $X$ and $Y .$

(a) Verify that
$$f(x, y)=\left\{\begin{array}{ll}{4 x y} & {\text { if } 0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1} \\ {0} & {\text { otherwise }}\end{array}\right.$$
is a joint density function.
(b) If $X$ and $Y$ are random variables whose joint density func-
tion is the function $f$ in part (a), find
(i) $P\left(X \geqslant \frac{1}{2}\right) \quad$ (ii) $P\left(X \geqslant \frac{1}{2}, Y \leqslant \frac{1}{2}\right)$
(c) Find the expected values of $X$ and $Y .$

Key Concepts

Expected Value Calculation in Continuous Distribution

The expected value of a continuous random variable is calculated by integrating the product of the variable and its probability density function over its entire range. In the context of joint distributions, the expected value of each variable is computed using double integrals, which may involve the marginal density functions if the variables are considered separately.

Double Integration in Probability

Double integration is a key mathematical tool used to compute probabilities and expectations in joint distributions. By integrating the joint density function over specific regions of its support, one can determine the probability associated with events defined by inequalities in the random variables.

Marginal Density Functions

Marginal density functions are derived from a joint density function by integrating out one of the variables. They represent the probability distribution of each individual random variable and are essential for analyzing the behavior of each variable separately.

Joint Probability Density Function

A joint probability density function (pdf) describes the likelihood of two continuous random variables occurring simultaneously. It is used to capture the combined probabilistic behavior of the random variables, defining the probability over a two-dimensional space through a function that integrates to 1 over its support.

Verification of Density Functions

Verifying that a function is a valid pdf involves checking that it is nonnegative over its entire domain and that the double integral of the function over the complete support equals 1. This ensures that the total probability is properly normalized for the random variables.

Transcript

00:01 F of x y is a joint density function if you have this double integral to be equal to 1 so since f of s is equal to 0 if x y is greater than 1 or x y less than 0 we can write the double integral from negative infinity to infinity you have negative infinity to infinity of f of x y the x the y to be equal to the integral from zero to one you have zero to one f of x y the x d y the x d y so this would be equal to the double integral zero to one the function is for x y the x die y the y so 4 is a constant out so you have from 0 to 1 x d x and 0 to 1 y d y so this will be 4 you have x squared divided by 2 0 to 1 y squared divided by 2 0 to 1 so this will be 4 you have 1 divided by 2 1 divided by 2 which is equal to 1 so hence we have proved that this is a joint density function so hence proved so this is for the first part then for the second part b and we have p is greater than no equal to 1 divided by 2 this is equal to the double integral to 0 to 1 you have 4x y the y d y the x so this would be equal to 1 and 2 you have 1 so we have this differentiates with respect to y so you have four x y squared divided by two the interval from zero to one d x so this will be equal to the integral one by two one you have two two x d x and this will be equal to the integral here that is x squared and this is to 1 minus 1 divided by 4 which is equal to 3 divided by 4 then for the second part i i i so for i i i i you have the probability that our x is greater than or equal to 1 divided by 2 and now y is less than or equal to 1 divided by 2.

04:23 So we have the double integral 1 divided by 2 to 1.

04:29 Then from 0 to 1 divided by 2, we have 4xy, d, y, d, x.

04:38 So this will be equal to have 1 divided by 2, 1.

04:45 So this will be 2 x y squared the interval from 0 to 1 onto the x and this will be equal to the integral 1 divided by 2 1 we have 1 divided by 2 x d x so this then it's then it's equal to 1 divided by 4 x squared the interval from half to 1 and this will give us 1 divided by 4 minus 1 divided by 16 and this is equal to 3 divided by 16 so what it means is that what it means is that so this implies that's so this implies that the probability that our x is greater than or equals 1 divided by 2, it's equal to 3 divided by 4...

Please give Ace some feedback