Let X and Y be independent random variables, uniformly distributed on [0,1] . Let U=min{X,Y} and V=max{X,Y} .

Let a=E[UV] and b=E[V]

1. Find a
2. Find b
3. Find Cov(U,V) . You can give either a numerical answer or a symbolic expression involving a and b . (There are many possible equivalent answers.)

a=E[UV] = 1/4

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EV= 1-EU = 1/3
COV(U,V) = E(UV) - EUEV=1/36

To find the values of a and b, we can use the properties of the uniform distribution and the definitions of U and V.

1. Finding a:

The random variable U represents the minimum of X and Y. Since X and Y are independent and uniformly distributed between 0 and 1, the probability density function (PDF) of U can be obtained as follows:

f_U(u) = f_X(u) * f_Y(u)
= 1 * 1 = 1, for u in [0,1]

To find the expected value of UV (denoted as E[UV] or a), we need to evaluate the integral of the product of U and V over the given range:

a = E[UV] = āˆ«āˆ«(uv) * f_U(u) * f_V(v) du dv
= āˆ«āˆ«(uv) * 1 * 1 du dv
= āˆ«āˆ«(uv) du dv, over the range 0 ā‰¤ u ā‰¤ v ā‰¤ 1

Integrating over u first:

a = āˆ«v*[(1/2)*u^2]du = [(1/2)*v*(1/3)*u^3] evaluated from u=0 to u=v
a = (1/2)*v*(1/3)*v^3 - 0
a = (1/6)*v^4

Therefore, a = (1/6)*v^4.

2. Finding b:

The random variable V represents the maximum of X and Y. Similarly to U, the PDF of V can be obtained:

f_V(v) = f_X(v) * f_Y(v)
= 1 * 1 = 1, for v in [0,1]

To find the expected value of V (denoted as E[V] or b), we evaluate the integral of V over the given range:

b = E[V] = āˆ«v * f_V(v) dv
= āˆ«v * 1 dv
= (1/2) * v^2, evaluated from v=0 to v=1
= (1/2) * 1^2 - (1/2) * 0^2
= (1/2)

Therefore, b = (1/2).

3. Finding Cov(U, V):

The covariance of two random variables, Cov(U,V), can be calculated as the difference between the expected value of their product and the product of their expected values:

Cov(U, V) = E[UV] - E[U]E[V]
= a - E[U]E[V]
= (1/6)*v^4 - (1/6)*v^4 * (1/2)
= (1/6)*v^4 - (1/12)*v^4
= (1/12)*v^4

So, Cov(U, V) = (1/12)*v^4 (where v is the maximum among X and Y, with v in the range [0,1]).

To find the values of a, b, and Cov(U,V) for the given problem, we need to use some properties of independent random variables and the formulas for expected value and covariance.

1. To find a = E[UV]:
Since X and Y are independent random variables uniformly distributed on [0,1], we can express the joint distribution function of X and Y as f(x,y) = 1 for 0 ā‰¤ x,y ā‰¤ 1.

The value of U = min{X,Y} is less than or equal to x only if both X and Y are less than or equal to x. Therefore, the cumulative distribution function (CDF) of U is given by:
F_U(x) = P(U ā‰¤ x) = P(min{X,Y} ā‰¤ x) = P(X ā‰¤ x, Y ā‰¤ x) = P(X ā‰¤ x)P(Y ā‰¤ x) = x*x = x^2

By taking the derivative of the CDF of U, we can find its probability density function (PDF):
f_U(x) = d/dx * F_U(x) = d/dx * (x^2) = 2x

Similarly, the value of V = max{X,Y} is greater than or equal to y only if both X and Y are greater than or equal to y. Therefore, the CDF of V is given by:
F_V(y) = P(V ā‰¤ y) = P(max{X,Y} ā‰¤ y) = P(X ā‰¤ y, Y ā‰¤ y) = P(X ā‰¤ y)P(Y ā‰¤ y) = y*y = y^2

By taking the derivative of the CDF of V, we can find its PDF:
f_V(y) = d/dy * F_V(y) = d/dy * (y^2) = 2y

To find E[UV], we use the formula for expected value:
a = āˆ«āˆ« u*f(u,v) du dv, where f(u,v) is the joint PDF of U and V.

We know that U is less than or equal to V for all values of V. Thus, we have f(u,v) = 2u*2v = 4uv.

Now we can find a:
a = āˆ«āˆ« u * 4uv du dv
= 4 āˆ«āˆ« u^2v du dv
= 4 āˆ«[0,1] āˆ«[0,v] u^2v du dv
= 4 āˆ«[0,1] [(1/3)*v^3] dv
= 4 * (1/3) * āˆ«[0,1] v^3 dv
= 4 * (1/3) * [v^4/4] [0,1]
= (1/3)

Therefore, the value of a is 1/3.

2. To find b = E[V]:
For a uniform distribution on [0,1], we can use the formula for expected value to find b:
b = āˆ« v * f(v) dv, where f(v) is the PDF of V.

Using the PDF we found for V, we can calculate b:
b = āˆ« v * 2v dv
= 2 āˆ« v^2 dv
= 2 * (1/3) * v^3 + C
= (2/3) * (v^3) + C

The limits of integration for v are 0 and 1, so substituting these values:
b = (2/3) * (1^3) + C - [(2/3) * (0^3) + C]
= (2/3) + C - (0 + C)
= (2/3)

Therefore, the value of b is 2/3.

3. To find Cov(U,V):
Covariance (Cov) can be calculated using the formula:
Cov(U,V) = E[UV] - E[U]E[V]

We already know the values of E[UV] (a) and E[V] (b), so we can plug them into the formula:
Cov(U,V) = a - E[U]E[V]
= a - (E[U])(b)
= (1/3) - (1/3) * (2/3)
= (1/3) - (2/9)
= 3/9 - 2/9
= 1/9

Therefore, the covariance of U and V is 1/9.