Probability Question - Change of Variables?

Question

Y~unif(0,1). Find a function phi(Y) such that X=phi(Y) has density f given by f(x) = 2x, whenever x is in [0,1], and f(x)=0 elsewhere.

Y~unif(0,1). Find a function phi(Y) such that X=phi(Y) has density f given by f(x) = 2x, whenever x is in [0,1], and f(x)=0 elsewhere.

I did:
phi(y)=P(Y =...
=...
=P(...)
=x^2, since phi'(y)=f(x)=2x.

So how do I know what phi(Y) might be?

Answer 1

Aha. There is a cute trick for this, but it is a standard one, in many texts.

Check out:
http://en.wikipedia.org/wiki/Inverse_transform_sampling

In this case, the cumulative distribution function that corresponds to f is monotonic and easily computed. Ditto for its inverse, so this approach works perfectly.

Recall that the cumulative distribution function F(x) is defined by:

F(x) = integral from -infinity to x of f(t) dt

where f(t) is the density function.

In this case, f(t) is 0 for t < 0 so the integral can start at 0 instead of -infinity.

Similarly, there is no point integrating beyond 1.

So F(x) is the integral of 2x dx from 0 to 1.

And your phi(Y) is just the inverse function of F. That is, phi is the function such that phi(F(x)) = x for all x.

(For example, if F(x) were sqrt(x), phi(t) would be t^2.)

Answer 2

With reposition: P(b1&r2) = P(b1)*P(r2) =(6/11)*5/11) = 30/121 P (r1&r2) = P(r1)*P(r2) = (5/11)*(5/11) = 25/121 with out reposition: P(b1&r2) = P(b1)*P(r2/b1) = (6/11)*(5/10) = 30/one hundred ten P (r1&r2) = P(r1)*P(r2/r1) = (5/11)*(4/10) = 20/one hundred ten discover the threat that the 2d disc bumped off is purple. With reposition: 30/121 + 25/121 = fifty 5/121 with out reposition: 30/one hundred ten + 20/one hundred ten = 50/one hundred ten