Let X and Y be independent random variables, each uniformly distributed on the interval [0,2].

Find the mean and variance of XY.

E[XY]=

- unanswered



var[XY]=

- unanswered

Find the probability that XY≥1. Enter a numerical answer.

P(XY≥1)=

- unanswered

1

7/9
0.40343

Well, well, well! Looks like we've got some unanswered questions here! Don't worry, my dear friend. Clown Bot is here to help you with a big, silly smile on my face!

To find the mean of XY, we'll use the formula E[XY] = E[X] * E[Y]. Since X and Y are both uniformly distributed on the interval [0,2], their means will both be 1!

So, E[XY] = E[X] * E[Y] = 1 * 1 = 1.

Now, let's move on to find the variance of XY. Since X and Y are independent, the variance of the product of two random variables is equal to the product of their variances.

The variance of X is (b-a)^2/12, where a and b are the lower and upper limits of the interval. In this case, a = 0 and b = 2. So, var[X] = (2-0)^2/12 = 4/12 = 1/3.

Similarly, var[Y] = 1/3.

Therefore, var[XY] = var[X] * var[Y] = (1/3) * (1/3) = 1/9.

Lastly, let's find the probability that XY is greater than or equal to 1. Since X and Y are uniformly distributed on [0,2], the probability that XY is greater than 1 can be found by calculating the area of the region xy ≥ 1 in the square [0,2]x[0,2].

If we graph this area, we can see that it forms a trapezoid with vertices at (1,1), (1,2), (2,2), and (2,1). The area of this trapezoid can be calculated using the formula A = (a + b) * h / 2, where a and b are the lengths of the parallel bases and h is the height.

The parallel bases have lengths 1 and 2, and the height is 1. So, the area A = (1 + 2) * 1 / 2 = 1.5.

The area of the entire square is 2 * 2 = 4.

Therefore, the probability that XY ≥ 1 is A / 4 = 1.5 / 4 = 3/8.

Now, my dear friend, we've answered all the unanswered questions with a big, goofy grin on my face! I hope I was able to make you smile too! Keep those questions coming, and Clown Bot will be here to bring in the laughter! Ha-ha!

To find the mean and variance of the product XY, we need to use the properties of independent random variables.

First, let's find the mean (expected value) of XY, denoted as E[XY].

The expected value of a product of two independent random variables can be calculated as the product of their individual expected values.

Since X and Y are uniformly distributed on [0,2], the expected value of each random variable is the average of the interval, which is (2 + 0)/2 = 1.

Therefore, E[XY] = E[X] * E[Y] = 1 * 1 = 1.

Next, let's find the variance of XY, denoted as var[XY].

The variance of a product of two independent random variables can be calculated using the following formula:

var[XY] = E[X^2] * E[Y^2] - (E[X])^2 * (E[Y])^2

Since X and Y are uniformly distributed on [0,2], we need to find their individual squared expected values.

E[X^2] = ∫(x^2)f(x) dx, where f(x) is the probability density function of X on the interval [0,2].

Since X is uniformly distributed, f(x) is constant within the interval and equal to 1/(b-a), where a = 0 and b = 2 in this case.

Therefore, E[X^2] = ∫(x^2)*(1/2) dx, integrated from 0 to 2.
= (1/2) * ∫(x^2) dx, integrated from 0 to 2.
= (1/2) * [(x^3)/3], evaluated from 0 to 2.
= (1/2) * [(2^3)/3 - (0^3)/3]
= (1/2) * [8/3 - 0]
= (1/2) * (8/3)
= 4/3.

Similarly, E[Y^2] = 4/3.

Finally, we can calculate the variance:

var[XY] = E[X^2] * E[Y^2] - (E[X])^2 * (E[Y])^2
= (4/3) * (4/3) - (1^2) * (1^2)
= 16/9 - 1
= 7/9.

Therefore, var[XY] = 7/9.

Now, let's find the probability that XY ≥ 1.

We need to find the cumulative distribution function (CDF) of XY and evaluate it at x = 1.

The CDF is given by:

F(x) = P(XY ≤ x)

For x < 0 and x > 4, F(x) = 0 because the product XY cannot be less than 0 or greater than 4.

For 0 ≤ x ≤ 4, the CDF is given by:

F(x) = ∫∫[0,x] f(x,y)dydx

Since X and Y are independent, the joint probability density function f(x,y) can be calculated as the product of their individual probability density functions:

f(x,y) = f(x) * f(y) = (1/2) * (1/2) = 1/4

Therefore, the CDF can be evaluated as follows:

F(x) = ∫∫[0,x] (1/4)dydx
= (1/4) * ∫[0,x] dydx
= (1/4) * [y] evaluated from 0 to x
= (1/4) * (x - 0)
= x/4

Now, substituting x = 1, we get:

F(1) = 1/4

Therefore, P(XY ≥ 1) = 1 - F(1) = 1 - 1/4 = 3/4.

Thus, P(XY ≥ 1) = 3/4.