Let X be a positive continuous random variable with density fXpxq. Let Y " lnpXq. (a) Find the density fY pyq of Y in terms of the density of X. (b) If X is a uniform random variable on the interval p0, 5s, what is fY pyq?

X density = fXpxq and

Y" =InpXq

Now to find Y density FYpyq interms of the density of X we compare the density of X with Y"

fX = In

And PXq =pxq

Thus replacing x with y,

PXq = pyq

(a) Hence the density of Y is FYpyq

(b) at p0, fYpyq =fYp0q= 0

At 5s, FYpyq =5

a) lnf(x)dx =[ xlnf(x)- x] + C

b) 5(ln5pq-1)

Step-by-step explanation:

Given f(x) = pxq,

Y=ln(pxq)

(a). f(y) = ∫lnf(x)dx over range x

Using integration by parts

∫udv = uv - ∫vdu

lnf(x) = u, dv =1, > v = x

∫udv = lnf(x). x - x∫(1/pxq)pq.dx

∫udv = xlnf(x) - ∫dx

∫lnf(x)dx =[ xlnf(x)- x] + C

(b) ∫lnf(x)dx = [xln(pxq) - x], at range x= 0, and x= 5

= 5ln5pq - 5

=5(ln5pq-1)

a

step-by-step explanation:

when a number is between two straight lines it means the absolute value.

to find absolute value you need to know how many numbers is that number away from 0.

when finding absolute value the number will always come out positive

so absolute value of -3 is 3

absolute value of 3/2 is 3/2

absolute value of -2 is 2

absolute value of 3.5 is 3.5

absolute value of -1 is 1

now put these numbers in order:

a)   |-1|, |3/2|, |-2|, |-3|, |3.5|