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]= 1
var[XY]= ???
Find the probability that XY≥1. Enter a numerical answer.
P(XY≥1)= ???
please solve P(XY≥1)=
var(XY) = E[(XY)^2] - (E[XY]^2) = 7/9
please solve P(XY≥1)= 0.40343
To find the mean and variance of XY, we need to use the properties of expected value and variance.
Mean (Expected Value):
The expected value of a product of independent random variables is equal to the product of their individual expected values. Since X and Y are uniformly distributed on the interval [0,2], their individual expected values are given by:
E[X] = E[Y] = (b + a) / 2, where a and b are the lower and upper bounds of the interval.
For X and Y, a = 0 and b = 2, so E[X] = E[Y] = (2 + 0) / 2 = 1.
Therefore, E[XY] = E[X] * E[Y] = 1 * 1 = 1.
Now, let's find the variance:
Variance:
The variance of a product of independent random variables can be calculated using the following formula:
var[XY] = E[X^2] * E[Y^2] - (E[X])^2 * (E[Y])^2
To find E[X^2] and E[Y^2], we need to know the distribution of X and Y.
Since X and Y are uniformly distributed on the interval [0,2], we can determine that their distributions are continuous uniform distributions with parameters a = 0 and b = 2. The variance of a continuous uniform distribution is given by:
var[U(a,b)] = (b - a)^2 / 12
For X and Y, var[X] = var[Y] = (2 - 0)^2 / 12 = 1/3.
To calculate E[X^2] and E[Y^2]:
E[X^2] = var[X] + (E[X])^2 = 1/3 + 1^2 = 1/3 + 1 = 4/3,
E[Y^2] = var[Y] + (E[Y])^2 = 1/3 + 1^2 = 1/3 + 1 = 4/3.
Now, substituting the values into the formula for 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.
Next, let's calculate the probability that XY ≥ 1:
P(XY ≥ 1) can be expressed as the complementary probability of P(XY < 1).
Since X and Y are uniformly distributed on the interval [0,2], the joint probability density function (PDF) of X and Y is constant over the region defined by 0 ≤ X ≤ 2 and 0 ≤ Y ≤ 2. The joint distribution is a square with area 4.
To find P(XY < 1), we need to find the area of the region where XY < 1 within the square:
We can consider two cases:
Case 1: X < 1 and Y < 1.
The region where both X and Y are less than 1 forms a smaller square with side length 1 (1x1). The area of this region is 1.
Case 2: X > 1 and Y > 1.
The region where both X and Y are greater than 1 forms another smaller square with side length 1 (1x1). The area of this region is 1.
Therefore, P(XY < 1) = Area of region where XY < 1 / Area of the whole square
= (1 + 1) / 4
= 2 / 4
= 1 / 2.
Finally, the probability that XY ≥ 1 is the complementary probability:
P(XY ≥ 1) = 1 - P(XY < 1) = 1 - 1/2 = 1/2.
Therefore, P(XY ≥ 1) = 1/2.