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  • Suppose we have a random variable $X$ with probability distribution $f(X = x)$
  • Suppose the cumulative probability distribution of this function is $F(X \leq x) = 1 - f(x)$
  • Suppose now we define a new function called the "survival function" as $S(X) = 1 - F(X)$

Consider the Exponential Distribution $f(X = x) = \lambda e^{-\lambda x}$ : The survival function for the Exponential Distribution is given by: $S(X) = e^{ -\lambda * x}$.

My Question : Now, I am interested in showing that a survival function $S(X) = e^{ -\lambda * x}$ can ONLY be produced by a probability distribution function of the form $f(X = x) = \lambda e^{-\lambda x}$. Is it possible to prove this? Can we prove that a certain function of a specific probability distribution is unique?

I.e. There does not exist some probability distribution function $g(X = x)$, whose cumulative function $G(X \leq x) = 1 - f(x)$, whose survival function $S'(X) = 1 - G(X)$ ... would somehow equal to $S(X)$? In other words, there is exists only one such $f(x)$ which will allow you to end up with a specific $S(X)$

Thanks!

Note: I know that under certain conditions, some probability distribution functions are equal to each other (e.g. gamma distribution = exponential distribution under certain conditions). For example:

\begin{equation} \text{Gamma Distribution:} \quad f(x;\alpha,\theta) = \frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha} \end{equation}

Where:

  • $\alpha$ is the shape parameter
  • $\theta$ is the scale parameter
  • $\Gamma(\alpha)$ is the gamma function.

\begin{equation} \text{Exponential Distribution:} \quad f(x;\lambda) = \lambda e^{-\lambda x} \end{equation}

Where:

  • $\lambda$ is the rate parameter.

If $\alpha=1$ and $\theta=1/\lambda$ in the gamma distribution:

\begin{equation} f(x;1,1/\lambda) = \frac{x^{1-1}e^{-x/(1/\lambda)}}{\Gamma(1)(1/\lambda)^1} = \lambda e^{-\lambda x} \end{equation}

Which is exactly the exponential distribution. The gamma distribution with $\alpha=1$ and $\theta=1/\lambda$ is the same as the exponential distribution with rate $\lambda$.

This means that a Gamma Distribution with $\alpha=1$ and $\theta=1/\lambda$ would have the same survival function as an Exponential Distribution ... but a Gamma Distribution with $\alpha=1$ and $\theta=1/\lambda$ IS an Exponential Distribution. Therefore, they obviously will have the same survival function.

I am interested in knowing if it is possible to start with some completely different probability distribution that can not be re-formulated/re-written as a Exponential Distribution .... but still somehow possible to end up with the same survival function of the Exponential Distribution. Can this be mathematically proven that this is possible? Can this be mathematically proven that this is impossible?

References:

stats_noob
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    yes, this is just a consequence of $\pi$-$\lambda$ theorem on uniqueness of measure. – Andrew Nov 17 '23 at 22:24
  • @Andrew: thank you so much for your reply! as I am barely unfamiliar with the concepts of measures, could you please write a short explanation/proof as to why this is true (if you have time)? thank you so much! – stats_noob Nov 17 '23 at 22:35

1 Answers1

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We can prove that if two probability measures $P, Q$ on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ have the same CDFs, then $P = Q$. Written out fully, this means if $P((-\infty,x]) = Q((-\infty, x])$ for all $x \in \mathbb{R}$ then $P(A) = Q(A)$ for all Borel sets $A$. The proof uses the $\pi$-$\lambda$ theorem. Since the collection of intervals $(-\infty, x]$ with $x \in \mathbb{R}$ is a $\pi$-system that generates $\mathcal{B}(\mathbb{R})$ and the collection of Borel sets $A$ for which $P(A) = Q(A)$ is a $\lambda$-system, agreement on that $\pi$-system implies agreement on the entire $\mathcal{B}(\mathbb{R})$.

Mason
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