How to compute the CDF of this random variable?

Question

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!

Answer 1

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

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

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

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 $P(X \geq 0|\eta,h,a)=1-CDF_X(0)$.

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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:

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

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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 $a < 1$

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

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