0
$\begingroup$

Suppose $X, Y, Z$ all follow a Poisson distribution each with a distinct mean parameter $\lambda_x, \lambda_y, \lambda_z$ respectively. Additionally assume distribution is independent of the others.

How would I go about calculating $P(Y < X < Z)$? My first guess would be something like

$$\sum_{k=1}^\infty \sum_{j=1}^\infty \sum_{i=1}^j f(i, j,k)$$

But I'm not sure what function to be summing over. Would it just be the product of their PMF's?

$\endgroup$
8
  • 1
    $\begingroup$ Yes, it will be the product of their probability mass functions. However, rethink the bounds; that's not what the should be. $\endgroup$ – Graham Kemp Feb 23 at 7:30
  • $\begingroup$ Would it be $ \sum_{z=0}^\infty \left(\sum_{k=z}^{\infty} Pos(k; \lambda_i)\right) \left(\sum_{l=0}^z Pos(l; \lambda_j)\right)Pos(z; \lambda)$? $\endgroup$ – blueteethbass Feb 23 at 7:34
  • $\begingroup$ @GrahamKemp Oops, forgot to tag you! $\endgroup$ – blueteethbass Feb 23 at 16:41
  • $\begingroup$ Hard to say. You want to measure $Y<X<Z$ and the parameters are $\lambda_X, \lambda_Y, \lambda_Z$. Use them. @Julian $\endgroup$ – Graham Kemp Feb 23 at 22:43
  • $\begingroup$ Right. I'm just discretizing the definition for continuous functions, and I imagine the technique is similar $\endgroup$ – blueteethbass Feb 23 at 23:03
1
$\begingroup$

Let us use the convenient symbols $x,y,z$ as the bound variables for the series.

Due to the independence, the joint probability mass function will be the product of their probability mass functions. So the term for the series is: $$\def\pois{\operatorname{\cal Pois}}\pois(x;\lambda_{\small X})\pois(y;\lambda_{\small Y})\pois(z;\lambda_{\small Z})$$

The series needs to measure over the event of $Y{<}X{<}Z$, so the domain is: $$\{\langle x,y,z\rangle{\in}\Bbb N^3: 1{\leqslant}x, 0{\leqslant}y{\leqslant}x{-}1, x{+}1{\leqslant}z\}$$

Or such.

$$\displaystyle\mathsf P(Y{<}X{<}Z) ~=~ \sum_{x=1}^\infty\pois(x;\lambda_{\small X})\left(\sum_{y=0}^{x-1}\pois(y;\lambda_{\small Y})\right)\left(\sum_{z=x+1}^{\infty}\pois(z;\lambda_{\small Z})\right)$$

$\endgroup$

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.