Question

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)

Answer 1

`"F"(x) = { (1-e^(-lambdax), x>=0), (0, x<=0) :}`.

Explanation:

You are given

`"P"(X>x+y | X>x)="P"(X>y)`.

By conditional probability,

`"P"(X>x+y | X>x)=("P"(X>x+y nn X>x))/("P"(X>x))`,

If `X>x+y` and `x,y > 0` then `X>x`.

Then,

`"P"(X>x+y | X>x)=("P"(X>x+y))/("P"(X>x))`.

Substituting,

`"P"(X>x+y)="P"(X>x)"P"(X>y)`.

By the definition of the cumulative distribution function we know that
`"P"(X>x) = 1-"F"(x)`.

We can substitute this in and find that,

`(1-"F"(x+y))=(1-"F"(x))(1-"F"(y))`.

If we define `"g"(x) = 1 - "F"(x)` as your hint suggests then,

`"g"(x+y)="g"(x)"g"(y)`.

I will now find the solutions to this functional equation. Taking logarithms of both sides gives

`ln("g"(x+y)) = ln("g"(x)) + ln("g"(y))`.

Defining `"G"(x) = ln("g"(x))` means that `"G"(x)` satisfies the functional equation

`"G"(x+y) = "G"(x)+"G"(y)`.

To solve for the most general `"G"(x)` that satisfies this equation write `"G"(x)` as it's series `"G"(x) = sum_n a_n x^n`.

Then,

`sum_n a_n (x+y)^n = sum_n a_n (x^n+y^n)`
`sum_n a_n (sum_k a_k x^k y^(n-k) ) = sum_n a_n(x^n+y^n)`
(if anyone can format binomial coefficients properly let me know!)

Coefficients of `x` and `y` on either side must be the same. For `n>=2`, on the left hand side we have powers of `xy` whereas these do not exist on the right hand side. We conclude `n<=1`. I.e. `"G"(x) = c_o + c_1x`.

`c_0 + c_1(x+y) = 2c_0 + c_1(x+y)`.

Then `c_0=2c_0=> c_0=0` and we conclude `"G"(x) = c_1x.`
As `"G"(x) = ln("g"(x))`, this implies that `"g"(x) = e^(c_1x)`.

We defined `"g"(x) = 1 - "F"(x)`. Then,

`"F"(x) = 1-e^(c_1x)`.

We already specified the property `x>0` earlier. As `x \to \infty`, `"F"(x) \to 1`. This means that `c_1` must be a negative number. Defined `c_1 = - \lambda`, for some `lambda > 0`.

Then,

`"F"(x) = { (1-e^(-lambdax), x>=0), (0, x<=0) :}`.

This gives us a probability density function `"f"(x)` of

`"f"(x) = {(lambda e^(-lambdax), x>=0), (0, x<=0):}`.

This is called an exponential distribution.