# How to compute the CDF of this random variable?

I'm working on a game theory model of incomplete information, where players observe certain attributes via noisy signals. Specifically, one player has the opportunity to choose any value $\eta$ from the interval $[0, h]$ to transfer to Player 2. $h$, the maximum he may transfer, is drawn from the uniform distribution $U(0,1)$. Player 1 knows $h$, but Player 2 only knows its distribution, and rather than observe it directly, receives a noisy signal $S_h \sim U(h - \epsilon, h + \epsilon)$ about its value. Last, Player 2 also receives a noisy signal $S_\eta \sim U(\eta - \epsilon, \eta + \epsilon)$ about Player 1's choice.

$a$ is some (known) multiplier on the interval $[0,1]$. Both players know what $a$ is. $a$ defines the threshold that Player 2 will use to decide how to behave. For example, if $a = .75$, then Player 2 will be cooperative if Player 1 chooses $\eta$ of at least 75% of $h$. How can I find the CDF:

$$P(S_\eta \geq a \times S_h \mid \eta, h, a)$$

This is the probability that Player 2's signal about $\eta$ will be greater than what he believes the threshold is.

I've done some simulations to calculate the CDF empirically, so I know that it looks roughly like that of the normal distribution:

(Edit: for parameters $h = .8, a = .75, \epsilon = .01$):

But I'd really like to solve for it analytically. Does anyone have an idea about how I could do that?

Thanks!

• Is \epsilon a predefined constant? Feb 16, 2016 at 5:27
• Yes, it's predefined and fixed. Feb 16, 2016 at 6:20

If $\epsilon$ is a fixed constant, then the conditional probability you are looking at is the difference of 2 uniform random variables.

$S_\eta \sim U[\eta - \epsilon, \eta + \epsilon]$

$S_h \sim U[h-\epsilon, h + \epsilon]$

Therefore,

$X\sim S_\eta - aS_h \sim \text{convolution}(U[\eta - \epsilon, \eta + \epsilon], U[-ah-a\epsilon, -ah + a\epsilon])$ in density.

The density should be a trapezium (piece-wise linear) and the CDF should be a truncated piece-wise line or parabola.

(The support region containing all the mass should be of order $2(a+1)\epsilon$, which is the sum of the support lengths of each uniform variable. I see from your plot that your CDF spans a region of $\approx 0.62-0.58 \approx 0.04$. So I guess your $\epsilon \approx 0.01$.)

The probability your are interested in in $P(X \geq 0|\eta,h,a)=1-CDF_X(0)$.

Convolution of two general rectangular pulses:

Let

$X_1\sim p_1(x)=\left\{\begin{matrix}u, x\in [a,b]\\ 0, \text{otherwise}\end{matrix}\right.$

and

$X_2 \sim p_2(x)=\left\{\begin{matrix}v, x\in[c,d]\\0,\text{otherwise}\end{matrix}\right.$

Then, $X_3=X_1+X_2 \sim p_3(x)=p_1 \ast p_2 (x) = \int p_1(y) p_2(x-y) dy$

By inspecting at which values of $x$, $p_2(x-y)$ is non-zero, we end up with:

$p_3(x)=uv \int\mathbb{I}_{[a,b]}\mathbb{I}_{[x-d,x-c]}dy$, where $\mathbb{I}(S)$ is the indicator function over set $S$.

$\Rightarrow p_3(x)=uv \text{ length}([a,b] \cap [x-d,x-c])$

By inspection again, of the above, we can see that the support region of $X_3$ is $x \in [c+a, d+b]$.

Let $p = \min \{b-a,d-c \}$ and $q = \max \{b-a,d-c\}$. This is defined because the shorter of the two pulses will determine the length of the plateau of the final trapezium, $p_3(x)$.

We have, then:

$p_3(x)=\left\{\begin{matrix}uvp\left(\dfrac{x-a-c}{p} \right) x\in[a+c,a+c+p]\\ uvp, x\in [a+c+p, a+c+q]\\ uvp \left( \dfrac{b+d-x}{p} \right) , x \in [a+c+q, b+d]\end{matrix}\right.$

Sanity check:

The area under the curve of $p_3$ is two triangles plus a rectangle, which works out to $uvpq$. In the case of uniform probability distributions, we have the requirement that $uv=1/(pq)$, so that the total area under the trapezium is unity.

If you perform the piece-wise integration the CDF becomes:

$C_3(x)=\left\{\begin{matrix}\dfrac{1}{2}uv(x-a-c)^2, x\in [a+c,a+c+p]\\ \dfrac{1}{2}uvp^2+uvp(x-a-c-p), x\in [a+c+p, a+c+q]\\ uvpq-\dfrac{1}{2}uv(b+d-x)^2 , x \in [a+c+q, b+d]\end{matrix}\right.$

In your problem, we have the following values to plug in into $C_3(x)$:

$[a,b]=[\eta-\epsilon, \eta+\epsilon]$

$[c,d]=[-ah-a\epsilon, -ah+a\epsilon]$

$p=\min\{b-a, d-c\}=2a\epsilon$, if $a < 1$

$q=\max\{b-a, d-c\}=2\epsilon$, if $a < 1$

$u=1/(2\epsilon)=1/q$

$v=1/(2a\epsilon)=1/p$

• This is great, thank you! So what would the CDF be exactly? I found some examples of convolution of uniform distributions online, but they only show an example for two $U(0,1)$ variables, so I'm having trouble extending it to to $U(\eta - \epsilon, \eta + \epsilon)$ and $U(-ah -a\epsilon, -ah + a\epsilon)$. The fact that it will be a piece-wise parabola with an inflection point at $ah$ makes sense, but I'm having trouble finding the exact shape. Feb 16, 2016 at 18:48
• I'd also like to find $P(\eta \geq ah \mid S_\eta, S_h)$, would this be a simple extension of the math above or is there a separate approach? Thanks! Feb 16, 2016 at 18:57
• I have a general expression for the CDF. While it looks daunting, there are a lot of cancellations such as u=1/q and v=1/p and uvpq=1. All you have to do is substitute the values at the end in C_3(x). Feb 18, 2016 at 2:37
• Your last question in the comment should probably be on another thread. I think it might require the application of Bayes' formula. You would need to look at the joint distributions on each pair of variables (eta,h) and (Seta,Sh). You would need a prior p(eta,h). The final P(eta>=ah|Seta,Sh) would be the area under the surface of the conditional: integrate on (eta >= ah) p((eta,h)|(Seta,Sh)) deta dh. Feb 18, 2016 at 2:58
• Thanks, I've created a new thread here. I understand conceptually that I need to find the conditional density and integrate over it, but I'm hung up on how to find that conditional distribution. Any help would be much appreciated! Mar 4, 2016 at 19:51