Let X be uniformly distributed on (0,1)

[TomJerry]

Question:
Let X be uniformly distributed on (0,1). Show that Y=- $$\lambda$$^-1 1n(1-X) has an exponential distribution with parameter $$\lambda$$>0

 

[chiro]

Do you know what generating functions are?

 

[statdad]

let

$$F_Y(y) = P(Y \le y)$$

and use the definition of $Y$ to rewrite the inequality in terms of $X$.

 

[TomJerry]

Do you know what generating functions are?

Nope !

 

[chiro]

Here is how you can do this with a moment generating function.

Basically you can prove that a distribution is a particular one if both moment generating functions are of the same type.

So for example you've given Y = - 1/(lambda) ln(1 - X).

We are given that X is uniform on (0,1) so basically standard uniform.

The PDF of X is simply 1 with the domain (0,1).

The MGF of Y is given by M(t) = E[e^(Yt)] = Integral [0,1] 1 x e^(-{1/lambda} x ln{1-x} x t) dx = Integral [0,1] (1-x)^(-t/{lambda})

Using a substitution we get u = 1 - x, du = -dx which gives us the integral

M(t) = Integral [0,1] u^(-t/{lambda}) du.

This gives us M(t) = 1/{-t/lambda + 1} which corresponds to the MGF of the exponential distribution.

 

[mtaboga]

You can also use the standard formula for computing the density of a strictly increasing function. If X has density f_X(x) and Y=g(X) is strictly increasing, then Y has density f_Y(y)=f_X(g^-1(y))*dg^-1(y)/dy