4
$\begingroup$

I saw an interesting formula at:

https://stats.stackexchange.com/questions/264861/probability-of-gamma-greater-than-exponential

but I didn't know how to derive it. The question is:

Let $X,Y$ be independent r.v.s and $X\sim\mathrm{Gamma}(k_1,\theta_1)$, $Y\sim\mathrm{Gamma}(k_2,\theta_2)$, with $\mathbb{E}[X]=k_1\theta_1$.

How to derive

$$P[X>Y]=IB(\frac{\theta_1}{\theta_1+\theta_2};k_2,k_1),$$

where $IB(x;a,b)$ is the regularized incomplete beta function defined as

$$IB(x;a,b)=\frac{1}{B(a,b)}\int_0^xt^{a-1}(1-t)^{b-1}dt,$$

and $B(a,b)$ is the beta function.

$\endgroup$
0

1 Answer 1

7
$\begingroup$

https://en.wikipedia.org/wiki/Beta_distribution#Derived_from_other_distributions https://en.wikipedia.org/wiki/Gamma_distribution#Related_distributions

First of all, a well known, preliminary result used here is that if $X \sim \text{Gamma}(k_1, \theta), Y \sim \text{Gamma}(k_2, \theta)$ and they are independent, then $$ \frac {X} {X+Y} \sim \text{Beta}(k_1, k_2)$$

Note that this result require both $X, Y$ have the same scale parameters, which you do not have yet. And as $\theta_1, \theta_2$ are the scale parameters, we have

$$ \frac {X} {\theta_1} \sim \text{Gamma}(k_1, 1), \frac {Y} {\theta_2} \sim \text{Gamma}(k_2, 1)$$

Therefore,

$$ \begin{align} \Pr\{X > Y\} &= \Pr\left\{\frac {X} {\theta_1} > \left(\frac {1} {\theta_1} + \frac {1} {\theta_2}\right) Y - \frac {Y} {\theta_2}\right\} \\ &= \Pr\left\{ \frac {Y} {\displaystyle \frac {Y} {\theta_2} + \frac {X} {\theta_1}} < \frac {\theta_1\theta_2} {\theta_1 + \theta_2} \right\} \\ &= \Pr\left\{ \frac {\displaystyle \frac {Y} {\theta_2}} {\displaystyle \frac {X} {\theta_1} + \frac {Y} {\theta_2} } < \frac {\theta_1} {\theta_1 + \theta_2} \right\} \\ &= F_Z\left(\frac {\theta_1} {\theta_1 + \theta_2} \right) \end{align}$$ where $$ Z = \frac {\displaystyle \frac {Y} {\theta_2}} {\displaystyle \frac {X} {\theta_1} + \frac {Y} {\theta_2} } \sim \text{Beta}(k_2, k_1)$$ and $F_Z$ is the CDF of $Z$. Finally you just need to use the result from Beta CDF:

https://en.wikipedia.org/wiki/Beta_distribution#Cumulative_distribution_function

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.