3
$\begingroup$

This question emerges from a project in microeconomic modeling.

I have $n$ agents receiving noisy i.i.d signals $s$.

In my model, a situation of interest occurs when the average signal across $n$ agents, say $\bar{s} := (1/n) \sum_{i=1}^n s$, passes a certain threshold $t$.

I would like to be able to express the probability that this happens, i.e. $P(\bar{s} > t)$, in an algebraically simple way.

Of course, $P(\bar{s} \geq t) = 1 - P(\bar{s} \leq t)$ so the probability I am interested in is a simple function of the CDF of $\bar{s}$.

My problem is I am unable to think of a distribution for $s$ that would make the CDF of $\bar{s}$ easy to play with and obtain analytical results from.

For example, letting $s \sim N(0,1)$ leaves me with a $P(\bar{s} \leq t)$ depending on the very unhandy error function (I know it's easy to obtain computation results based on $erf(\cdot)$ but I am aiming at analytical results here). Letting $s$ be a Bernoulli or a Uniform leaves me with the CDF of a Binomial or an Irwin-Hall, which aren't much nicer to play with...

Any suggestions as to what I could use for the distribution of $s$?

Bonus requirements:

  • in another part of my model, I am also interested in $P({s} \geq t)$ itself, so it would help a ton if both $s$ and $\bar{s}$ had simple CDFs.
  • for reasons I cannot yet lay down myself, I feel like I will be happier further down the line if $s$ has a continuous and somewhat symmetric distribution, but I am happy to hear about discrete and asymmetric solutions as well.
$\endgroup$
  • $\begingroup$ Is $n$ large? If so, can use the central limit theorem $\endgroup$ – user17900 Mar 9 at 16:59
  • $\begingroup$ No, unfortunately, I cannot afford to assume that 𝑛 is particularly large. Besides, even then, wouldn't relying on the central limit theorem still give me a CDF with an ugly π‘’π‘Ÿπ‘“(β‹…) in it? $\endgroup$ – Martin Van der Linden Mar 9 at 17:51
  • 1
    $\begingroup$ Tough for me to see how it could get any easier than Bernoulli or Uniform. Is there anyway to change the model? Perhaps so that you need the nth order statistic to cross a threshold? Sorry this isn't more helpful $\endgroup$ – Mmmmmm Mar 9 at 20:10
  • 1
    $\begingroup$ My first thought is that the exponential distribution may give you cleaner results. I will muck around with it for a bit. $\endgroup$ – Kitsune Cavalry Mar 9 at 20:23
  • 1
    $\begingroup$ @Mmmmmm: Thanks for the suggestion. I might consider that as a last resort. Maybe I'll eventually have to look at the median instead of the mean of the signals. But for the moment I would like to stick to the mean and keep exploring my options. Kitsune: Thanks, look forward to reading from you. $\endgroup$ – Martin Van der Linden Mar 9 at 20:43
2
$\begingroup$

So we have

$S_n \thicksim^{iid} \ ?$

$\bar{s} = \frac{1}{n}\sum{s_n}$

Exponential

Suppose $S_n$ follows the exponential distribution.

$$f(s|\beta) = \frac{1}{\beta} e^{-\frac{s}{\beta}} \quad , \quad 0 \leq s < \infty \quad , \quad \beta > 0$$

Take the simple bivariate case. Say $Z = \frac{S_1 + S_2}{2}$ and $W = S_1$.

So $S_2 = 2Z - W$ and $S_1 = W$. The determinant of the Jacobian: $\frac{\partial s_1}{\partial z} \frac{\partial y}{\partial w} - \frac{\partial x}{\partial w} \frac{\partial y}{\partial z} = -2$. The absolute value is $2$.

Thus, given the join marginal of $S_1 , S_2$

$$f(S_1, S_2) = \frac{1}{\beta ^2} e^{-\frac{S_1 + S_2}{\beta}}$$

we then find the joint marginal of $W, Z$, and integrate out $W$ to find the density function for $\frac{S_1 + S_2}{2}$

$$ \begin{align} h(W,Z) = & \ f(W, 2Z-W) \cdot 2 \\ = & \ \frac{2}{\beta ^2} e^{-\frac{w+2z-w}{\beta}} \\ = & \ \frac{2}{\beta ^2} e^{-\frac{2z}{\beta}} \\ \end{align} $$

Integrate with respect to $w$ from $z$ to $0$ gets you the density:

$$f(Z) = \frac{2}{\beta ^2} z e^{-\frac{2z}{\beta}}$$

Then you do a lot of garbage integration by parts (let the buyer beware if I have made an error) to get the cumulative distribution function evaluated from $z$ to $0$:

$$\boxed{F(Z) = -\frac{2z + \beta}{2 \beta} e^{-\frac{2z}{\beta}} + \frac{1}{2}}$$

Substitute $q = -\frac{2z}{2\beta}$ and you can make this look more tolerable:

$$\boxed{F(Z) = \left(q - \frac{1}{2}\right) e^{-2q} + \frac{1}{2}}$$

You may now umm...generalize the case for $n$ agents LOL. It may be more viable to compare this to the CDF of a single exponential distribution, then maybe try the case for $n=3$ and see if you can see a pattern.


Cauchy :P

One thing I looked at for fun online, take a simplified Cauchy distribution:

$$f(x) = \frac{1}{\pi(1+x^2)}$$

By the convolution formula and then a lot of work, the density of $Z = X + Y$ is:

$$\frac{2}{\pi(4+z^2)}$$

You can also verify that if you want the density of $Z = \frac{X + Y}{2}$, it is:

$$f(z) = \frac{1}{\pi(1+z^2)}$$

It's the Cauchy again! Now, the Cauchy has that pesky property where umm...it has no moments. But I'm pretty sure for this Cauchy distribution, the density is symmetric around 0. And technically you did not specify that your signals would have mean and variance, just that they were iid. :)

The cdf of this ends up being:

$$F(z) = \frac{\arctan(z)}{\pi} + \frac{1}{2}$$

So if the average for three variables or four variables also gives the density of the Cauchy, you may feel so bold as to use this.


Final Comments

Uhh, if you are willing discretize the noisy signal (you mentioned Bernoulli), you can try using the Poisson distribution or something. If you are looking into another continuous option you could try Beta. If I have more time I may edit this answer with some investigation into one or the other.

| improve this answer | |
$\endgroup$
  • $\begingroup$ The sum of $n$ mutually independent exponentials is now to follow an Erlang($\lambda$,$n$) (see math.wm.edu/~leemis/chart/UDR/PDFs/ExponentialErlang.pdf for example). I am afraid its CDF will still be too complex to play with but it might be my best bet, especially if I decide to look at low values of $n$. Thanks for the Cauchy suggestion too. As you wrote I did not say anything about needing moment ;) although, as you guessed, it's likely that I will further down the line. Btw, the sum of independent Poisson's is a Poisson, but again the CDF of a Poisson is not all that friendly. $\endgroup$ – Martin Van der Linden Mar 10 at 12:41
  • $\begingroup$ Very useful, thanks for the additional info. And yea I figured n exponentials would probably get messy. I think whatever you pick you are going to have some difficulties. $\endgroup$ – Kitsune Cavalry Mar 10 at 17:49

Your Answer

By clicking β€œPost 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.