Let F(x) be the cdf of the continuous-type random variable X, and assume that F(x)=0 for x<=0 and 0<F(x),1 for 0<x. Prove that if P(X>x+y|X>x)=P(X>y), then F(x)=1-#e^(-lamdax# , 0<x ?
Hint: show that g(x)=1-F(x) satisfies the functional equation
g(x+y)=g(x)g(y),
which implies that g(x)=a^(cx)
Hint: show that g(x)=1-F(x) satisfies the functional equation
g(x+y)=g(x)g(y),
which implies that g(x)=a^(cx)
1 Answer
Explanation:
You are given
By conditional probability,
If
Then,
Substituting,
By the definition of the cumulative distribution function we know that
We can substitute this in and find that,
If we define
I will now find the solutions to this functional equation. Taking logarithms of both sides gives
Defining
To solve for the most general
Then,
(if anyone can format binomial coefficients properly let me know!)
Coefficients of
Then
As
We defined
We already specified the property
Then,
This gives us a probability density function
This is called an exponential distribution.