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 $h = .8, a = .75, \epsilon = .01$): [1]

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


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

1 Answer 1


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

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

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


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


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


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

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 $a < 1$



  • $\begingroup$ 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. $\endgroup$
    – sundance
    Feb 16, 2016 at 18:48
  • $\begingroup$ 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! $\endgroup$
    – sundance
    Feb 16, 2016 at 18:57
  • $\begingroup$ 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). $\endgroup$ Feb 18, 2016 at 2:37
  • $\begingroup$ 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. $\endgroup$ Feb 18, 2016 at 2:58
  • $\begingroup$ 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! $\endgroup$
    – sundance
    Mar 4, 2016 at 19:51

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