Question

In: Statistics and Probability

Let Ul , U2 , U3 , U4 , U5 be independent, each with uniform distribution...

Let Ul , U2 , U3 , U4 , U5 be independent, each with uniform distribution on (0,1). Let R

be the distance between the minimum and the maximum of the Ui's. Find

a) E(R);

b) the joint density of the minimum and maximum of the U;'s;

c) P(R> 0.5)

Please do b) and c) and explain in details.

Solutions

Expert Solution

We are asked to find the minimum and maximum of the Ui's.

Let's take x as the maximum of Ui and y as the minimum of Ui.

Cumulative distribution for x will be : x5

density function will be : 5x4

Thus, for any value of x, we will get a distribution function for (x-y) : C(x-y)4

This will have value 1 when y=0, so Cx4=1 which leads to C= 1/x4

Now, for x=1, the distribution function for x will depend upon 4 variables and the maximum for 1-y will have a distribution function (1-y)4 and a density function 4(1-y)3

Now, our density function for (1-y) given x will be equal to 4(x-y)3 / x4

(This will also be the conditional density function for y given x)

Thus, our joint density function will be equal to : (5x4) (4 (x-y)3/x4)

(x4 can be cancelled from both numerator and denominator)

= 20 (x-y)3

Now, from the above equation, we can see that our joint density function will only depend on (x-y).

Now, let's take it as z. (i.e. z= x-y)

Now, for a uniform distribution of x & y, the probability density for z , (given x is at least y) will be proportional to 1-z.

( remember, our first choice of y can be anywhere from 0 to 1-z at which point, x is determined. Or, our first choice of x can be anywhere from z to 1 at which point y is determined. )

When moving from a joint density function for x & y to a single density function for z,

we get something proportional to (1-z) (20) (z3)

(20 can be removed, as we are looking at proportions)

Thus. it's simply proportional to (1-z)(z3)

Now, let's make it a proper function f(z) as,

f(z) = k (1-z) (z3)

= k ( z3-z4 )

integrating from 0 to z,

f(1) = (1/4 - 1/5) = 1/20 , So k=20, thus F(z)= 20 (z4/4 - z5/5) = 5z4 - 4z5

The probability that z will be less than or equal to 1/2 , F(1/2) = 5/16 - 4/32 = 5/16 - 2/16 = 3/16

Thus, P(R>0.5) = 1 - F(1/2) = 1 - 3/16 = 13/16   


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