Answered: We have a random variable X and Y that…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

We have a random variable X and Y that jave the joint pdf f(x) = {1 0<x<1, 0<y<1} {0 otherwise}

Let U = Y - X². What is the support for the random variable U? Are there critical points?

If U = Y/X. What is the support for the random variable U? Are there critical points?

Definition Definition Probability of occurrence of a continuous random variable within a specified range. When the value of a random variable, Y, is evaluated at a point Y=y, then the probability distribution function gives the probability that Y will take a value less than or equal to y. The probability distribution function formula for random Variable Y following the normal distribution is: F(y) = P (Y ≤ y) The value of probability distribution function for random variable lies between 0 and 1.

Expert Solution

Step 1

Sol:-

Part 1 :-

Let's first consider the transformation U = Y - X^2. We can find the support for U by determining the range of possible values that U can take.

To do this, we need to find the range of (X,Y) values that can produce each possible U value.

For any fixed value of U, we can solve for Y in terms of X using the expression U = Y - X^2. This gives us Y = U + X^2.

The support of U is the set of all U values for which there exists at least one (X,Y) pair that maps to that value of U.

Consider the range of possible (X,Y) values for which 0 < X < 1 and 0 < Y < 1.

For any given X value in this range, the smallest possible value of Y is X^2. This is because Y is bounded below by 0 and U = Y - X^2 cannot be negative.

For any given X value in this range, the largest possible value of Y is 1. This is because the joint pdf f(X,Y) is 0 for any (X,Y) pair outside the unit square.

Therefore, for any given value of U, we can find the corresponding range of X values that can produce that value of U. Specifically, we have:

U < 0: This can be achieved if and only if X^2 > 1 - U, which implies that |X| > sqrt(1 - U).

Therefore, the support for U in this case is:

-sqrt(1 - U) < X < sqrt(1 - U)

U >= 0: This can be achieved if and only if X^2 < 1 - U. Therefore, the support for U in this case is:

-sqrt(1 - U) < X < sqrt(1 - U)

Thus, the support of U is:

-sqrt(1 - U) < X < sqrt(1 - U)

And there are no critical points because U is a continuous random variable.