<p>So we have</p>

<p><span class="math-container">$S_n \thicksim^{iid} \ ?$</span></p>

<p><span class="math-container">$\bar{s} = \frac{1}{n}\sum{s_n}$</span></p>

<p><strong>Exponential</strong></p>

<p>Suppose <span class="math-container">$S_n$</span> follows the exponential distribution.</p>

<p><span class="math-container">$$f(s|\beta) = \frac{1}{\beta} e^{-\frac{s}{\beta}} \quad , \quad 0 \leq s &lt; \infty \quad , \quad \beta &gt; 0$$</span></p>

<p>Take the simple bivariate case. Say <span class="math-container">$Z = \frac{S_1 + S_2}{2}$</span> and <span class="math-container">$W = S_1$</span>.</p>

<p>So <span class="math-container">$S_2 = 2Z - W$</span> and <span class="math-container">$S_1 = W$</span>. The determinant of the Jacobian: <span class="math-container">$\frac{\partial s_1}{\partial z} \frac{\partial y}{\partial w} - \frac{\partial x}{\partial w} \frac{\partial y}{\partial z} = -2$</span>. The absolute value is <span class="math-container">$2$</span>.</p>

<p>Thus, given the join marginal of <span class="math-container">$S_1 , S_2$</span></p>

<p><span class="math-container">$$f(S_1, S_2) = \frac{1}{\beta ^2} e^{-\frac{S_1 + S_2}{\beta}}$$</span></p>

<p>we then find the joint marginal of <span class="math-container">$W, Z$</span>, and integrate out <span class="math-container">$W$</span> to find the density function for <span class="math-container">$\frac{S_1 + S_2}{2}$</span></p>

<p><span class="math-container">$$
\begin{align}
h(W,Z) = &amp; \ f(W, 2Z-W) \cdot 2 \\
= &amp; \ \frac{2}{\beta ^2} e^{-\frac{w+2z-w}{\beta}} \\
= &amp; \ \frac{2}{\beta ^2} e^{-\frac{2z}{\beta}} \\
\end{align}
$$</span></p>

<p>Integrate with respect to <span class="math-container">$w$</span> from <span class="math-container">$z$</span> to <span class="math-container">$0$</span> gets you the density:</p>

<p><span class="math-container">$$f(Z) = \frac{2}{\beta ^2} z e^{-\frac{2z}{\beta}}$$</span></p>

<p>Then you do a lot of garbage integration by parts (<em>let the buyer beware if I have made an error</em>) to get the cumulative distribution function evaluated from <span class="math-container">$z$</span> to <span class="math-container">$0$</span>:</p>

<p><span class="math-container">$$\boxed{F(Z) = -\frac{2z + \beta}{2 \beta} e^{-\frac{2z}{\beta}} + \frac{1}{2}}$$</span></p>

<p>Substitute <span class="math-container">$q = -\frac{2z}{2\beta}$</span> and you can make this look more tolerable:</p>

<p><span class="math-container">$$\boxed{F(Z) = \left(q - \frac{1}{2}\right) e^{-2q} + \frac{1}{2}}$$</span></p>

<p>You may now umm...generalize the case for <span class="math-container">$n$</span> agents LOL. It may be more viable to compare this to the CDF of a single exponential distribution, then maybe try the case for <span class="math-container">$n=3$</span> and see if you can see a pattern.</p>

<hr>

<p><strong>Cauchy</strong> :P</p>

<p>One thing I looked at for fun online, take a simplified Cauchy distribution:</p>

<p><span class="math-container">$$f(x) = \frac{1}{\pi(1+x^2)}$$</span></p>

<p>By the convolution formula and then a lot of work, the density of <span class="math-container">$Z = X + Y$</span> is:</p>

<p><span class="math-container">$$\frac{2}{\pi(4+z^2)}$$</span></p>

<p>You can also verify that if you want the density of <span class="math-container">$Z = \frac{X + Y}{2}$</span>, it is:</p>

<p><span class="math-container">$$f(z) = \frac{1}{\pi(1+z^2)}$$</span></p>

<p>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. :)</p>

<p>The cdf of this ends up being:</p>

<p><span class="math-container">$$F(z) = \frac{\arctan(z)}{\pi} + \frac{1}{2}$$</span></p>

<p>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.</p>

<hr>

<p><strong>Final Comments</strong></p>

<p>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.</p>

CC BY-SA 4.0
2020-03-10T12:41:36.953Z
The sum of $n$ mutually independent exponentials is now to follow an Erlang($\lambda$,$n$) (see http://www.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.
Martin Van der Linden
CC BY-SA 4.0
2020-03-10T17:49:25.957Z
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.
Kitsune Cavalry
CC BY-SA 4.0
2020-03-09T22:31:13.060Z
2020-03-09T22:31:13.060Z
Kitsune Cavalry
<p>This question emerges from a project in microeconomic modeling.</p>

<p>I have <span class="math-container">$n$</span> agents receiving noisy i.i.d signals <span class="math-container">$s$</span>.</p>

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

<p>I would like to be able to express the probability that this happens, i.e. <span class="math-container">$P(\bar{s} &gt; t)$</span>, in an algebraically simple way.</p>

<p>Of course, <span class="math-container">$P(\bar{s} \geq t) = 1 - P(\bar{s} \leq t)$</span> so the probability I am interested in is a simple function of the CDF of <span class="math-container">$\bar{s}$</span>.</p>

<p>My problem is I am unable to think of a distribution for <span class="math-container">$s$</span> that would make the CDF of <span class="math-container">$\bar{s}$</span> easy to play with and obtain <em>analytical</em> results from.</p>

<p>For example, letting <span class="math-container">$s \sim N(0,1)$</span> leaves me with a <span class="math-container">$P(\bar{s} \leq t)$</span> depending on the very unhandy error function (I know it's easy to obtain computation results based on <span class="math-container">$erf(\cdot)$</span> but I am aiming at <em>analytical</em> results here). Letting <span class="math-container">$s$</span> 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...</p>

<p>Any suggestions as to what I could use for the distribution of <span class="math-container">$s$</span>?</p>

<p><strong>Bonus requirements</strong>: </p>

<ul>
<li>in another part of my model, I am also interested in <span class="math-container">$P({s} \geq t)$</span> itself, so it would help a ton if <em>both</em> <span class="math-container">$s$</span> and <span class="math-container">$\bar{s}$</span> had simple CDFs.</li>
<li>for reasons I cannot yet lay down myself, I feel like I will be happier further down the line if <span class="math-container">$s$</span> has a continuous and somewhat symmetric distribution, but I am happy to hear about discrete and asymmetric solutions as well.</li>
</ul>

CC BY-SA 4.0
2020-03-09T17:51:21.400Z
No, unfortunately, I cannot afford to assume that  is particularly large. Besides, even then, wouldn't relying on the central limit theorem still give me a CDF with an ugly (⋅) in it?
Martin Van der Linden
CC BY-SA 4.0
2020-03-09T20:23:18.460Z
My first thought is that the exponential distribution may give you cleaner results. I will muck around with it for a bit.
Kitsune Cavalry
CC BY-SA 4.0
2020-03-09T20:43:28.020Z
@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.
Martin Van der Linden
CC BY-SA 4.0
2020-03-10T12:34:36.283Z
@KennyLJ: You may be right. Don't know if it's an indicator of what would happen at Stats.SE but I tried to ask a similar question on Math.SE and it wasn't received well. The question is not purely mathematical. Economists have a visceral understanding of the need to keep the pieces of a model "simple" in order to get analytical results. People at Math.SE got annoyed by the fact that "simplicity" is not formally defined and answered things like: "any function that is nondecreasing, right-continuous, has a limit of 1 as →∞, has a limit of 0 as →−∞: Choose any simple one that you like".
Martin Van der Linden
<microeconomics><probability>
Modelling involving sum of random variables: Simple CDF?
Martin Van der Linden
# Modelling involving sum of random variables: Simple CDF?

(Asked by: username_0 on 2020-03-09)

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.

**Tags:** microeconomics, probability

> **Comments:**
> **username_0**: No, unfortunately, I cannot afford to assume that  is particularly large. Besides, even then, wouldn't relying on the central limit theorem still give me a CDF with an ugly (⋅) in it?  
> **username_1**: My first thought is that the exponential distribution may give you cleaner results. I will muck around with it for a bit.  
> **username_0**: @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.  
> **username_0**: @KennyLJ: You may be right. Don't know if it's an indicator of what would happen at Stats.SE but I tried to ask a similar question on Math.SE and it wasn't received well. The question is not purely mathematical. Economists have a visceral understanding of the need to keep the pieces of a model "simple" in order to get analytical results. People at Math.SE got annoyed by the fact that "simplicity" is not formally defined and answered things like: "any function that is nondecreasing, right-continuous, has a limit of 1 as →∞, has a limit of 0 as →−∞: Choose any simple one that you like".  

---

## 1 answers

### Answer by username_1 on 2020-03-09. Score: 1

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.

> **Comments:**
> **username_0**: The sum of $n$ mutually independent exponentials is now to follow an Erlang($\lambda$,$n$) (see http://www.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.  
> **username_1**: 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.  

---

