Let X be uniformly distributed on (0,1)

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The discussion focuses on proving that the transformation Y = -λ⁻¹ ln(1-X), where X is uniformly distributed on (0,1), results in Y having an exponential distribution with parameter λ > 0. The moment generating function (MGF) approach is highlighted, demonstrating that if two distributions have the same MGF, they are of the same type. The MGF of Y is derived through integration, leading to a form that matches the MGF of the exponential distribution. Additionally, the density function transformation method is mentioned as an alternative proof. Overall, the discussion effectively establishes the relationship between the uniform and exponential distributions through mathematical transformations.
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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
 
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Do you know what generating functions are?
 
let

<br /> F_Y(y) = P(Y \le y)<br />

and use the definition of Y to rewrite the inequality in terms of X.
 
chiro said:
Do you know what generating functions are?

Nope !
 
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.
 
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
 
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