Survival in two period game: mean of z|z

Question

I am looking for the functional form to describe the following:

A random shock $x\sim Uniform(a,b)$ is multiplied with a second shock $y\sim Uniform(c,d)$.

What is the mean value of all combined shocks $z=xy$ below a certain threshold $v$?

Or to put it differently: $\mathbb{E}(z\vert z<v$).

For concreteness let's assume the following values:

$x\sim Uniform(0.5,1.5)$, $y\sim Uniform(0.4,1.6)$, and $v=0.7$.

To give you some background why I am asking this, I am trying to model the following:

In a two period game, each period a shock arrives and each period there is a cut off point determining who continues the game and who leaves (truncation). Those continuing receive the future shock (multiplication of distributions) and at the end we decide again how many are above and below the threshold. Now we want to know the average value of those who remain in the end (below threshold $v$).

This question is closely related to:

CDF of Z=XY with X~Uniform(0.5,1.5) and Y~Uniform(0.8,1.5)

but for the truncation and the mean part.

Any hints are welcome.

Answer 1

The details are tedious (some are shown), but at least we can look at the concrete example. Let $a=0.5$, $b=1.5$, $c=0.4$, $d=1.6$, and $k=(b-a)(d-c).$

Following the development shown in CDF of Z=XY with X~Uniform(0.5,1.5) and Y~Uniform(0.8,1.5) that the OP mentioned, we can find the CDF:

$$ F_Z(z)= \begin{cases} \left( \frac{1}{k} \right) \left[ac-z+z \left( \mathrm{ln} \ z - \mathrm{ln} \ a - \mathrm{ln} \ c \right) \right] \ , & 0.2 \le z \le 0.6 \\ \left( \frac{1}{k} \right) \left[ac-bc-z \ \mathrm{ln} \ a + z \ \mathrm{ln} \ b \right] \ , & 0.6 \le z \le 0.8 \\ \left( \frac{1}{k} \right) \left[ac-bc-ad+z -z \ \mathrm{ln} \ z + z \ \mathrm{ln} \ b + z \ \mathrm{ln} \ d \right] \ , & 0.8 \le z \le 2.4 \end{cases}$$

If we evaluate this at $z=0.7$ we find

$$q=P[Z \le 0.7]=\frac{c(a-b)+0.7(\mathrm{ln} \ b-\mathrm{ln} \ a)}{k} \approx 0.3075$$

By differentiating the cdf, we get the pdf of $z$ as

$$ f_Z(z)=\begin{cases} \left( \frac{1}{k} \right) \left[ \mathrm{ln} \ z - \mathrm{ln} \ a - \mathrm{ln} \ c \right] \ , & 0.2 \le z \le 0.6 \\ \left( \frac{1}{k} \right) \left[ \mathrm{ln} \ b - \mathrm{ln} \ a \right] \ , & 0.6 \le z \le 0.8 \\ \left( \frac{1}{k} \right) \left[ - \mathrm{ln} \ z + \mathrm{ln} \ b + \mathrm{ln} \ d \right] \ , & 0.8 \le z \le 2.4 \end{cases}$$

Now since we are interested in $Z \le 0.7,$ the conditional pdf will be

$$ f_{Z|Z \ \le \ 0.7}(z)=\begin{cases} \left( \frac{1}{kq} \right) \left[ \mathrm{ln} \ z - \mathrm{ln} \ a - \mathrm{ln} \ c \right] \ , & 0.2 \le z \le 0.6 \\ \left( \frac{1}{kq} \right) \left[ \mathrm{ln} \ b - \mathrm{ln} \ a \right] \ , & 0.6 \le z \le 0.7 \end{cases}$$

Now applying the definition of expected value, we have $$E[Z|Z \le 0.7] = \int z \ f_{Z|Z \ \le \ 0.7}(z) \ dz$$

Solving and evaluating the integrals leads to

$$E[Z|Z \le 0.7] = \frac{1}{100kq} \left[ 9 \ \mathrm{ln} (0.36)-\mathrm{ln} \ (0.04)-8-\frac{45 \ \mathrm{ln} \ a}{2} - 16 \ \mathrm{ln} \ c +\frac{13 \ \mathrm{ln} \ b}{2} \right] $$

$$$$

$$ E[Z|Z \le 0.7] \approx 0.5126$$

This is easily verified by simulation.

Answer 2

I will expand on soakley's answer to develop the formulas describing the conditional mean of the product of two uniform distributions.

The conditional mean is the integral of $z$ times the PDF from the lower boundary up to $z$ depending on the case and divided by the CDF at that point: $$ \mathbb{E}[Z|Z \le z]=\frac{\int^z z f_{Z}(z)dz}{F_Z(z)} $$.

This will yield the formulas I was looking for in this question: $$ \mathbb{E}[Z|Z \le z]= \begin{cases} \frac{- \frac{a^{2} c^{2}}{4} + \frac{z^{2}}{2} \log{\left (a \right )} - \frac{z^{2}}{2} \log{\left (\frac{z}{c} \right )} + \frac{z^{2}}{4}}{- a c + z \log{\left (a \right )} - z \log{\left (\frac{z}{c} \right )} + z} &, ac \le z \le bc \\ \frac{- \frac{a^{2} c^{2}}{4} + \frac{b^{2} c^{2}}{4} + \frac{z^{2}}{2} \log{\left (a \right )} - \frac{z^{2}}{2} \log{\left (b \right )}}{- a c + b c + z \log{\left (a \right )} - z \log{\left (b \right )}} &, bc \le z \le ad\\ \frac{\frac{a^{2} c^{2}}{4} - \frac{a^{2} d^{2}}{4} - \frac{b^{2} c^{2}}{4} + \frac{z^{2}}{2} \log{\left (b \right )} - \frac{z^{2}}{2} \log{\left (\frac{z}{d} \right )} + \frac{z^{2}}{4}}{a c - a d - b c + z \log{\left (b \right )} - z \log{\left (\frac{z}{d} \right )} + z} &, ad \le z \le bd \end{cases} $$

What helped me solve this was the sympy symbolic library. Find attached some code in python highlighting the steps:

from sympy import * 
from sympy.stats import *
init_printing()
x,y,z,u,v,a,b,c,d = symbols('x y z u v a b c d')
k = (b-a)*(d-c)
F_uv = integrate(z/x-v,(x,u,z/v))
F_uv

Derivation of first case:

F_Z = (F_uv.subs([(u,a),(v,c)]))/k
F_Z

f_Z = diff(F_Z,z)
f_Z

cond_F_Z = integrate(z*f_Z,(z,a*c,z))/F_Z
simplify(cond_F_Z)

N(cond_F_Z.subs([(a, .5), (b, 1.5), (c, .4), (d, 1.6), (z,.55 )]))

Derivation of second case:

F_Z2 = (F_uv.subs([(u,a),(v,c)])-F_uv.subs([(u,b),(v,c)]))/k
F_Z2

f_Z2 = diff(F_Z2,z)
f_Z2

cond_F_Z2 = (integrate(z*f_Z,(z,a*c,b*c))+integrate(z*f_Z2,(z,b*c,z)))/F_Z2
simplify(cond_F_Z2)

N(cond_F_Z2.subs([(a, .5), (b, 1.5), (c, .4), (d, 1.6), (z,.7 )]))

Derivation of third case:

F_Z3 = (F_uv.subs([(u,a),(v,c)])-F_uv.subs([(u,a),(v,d)])-F_uv.subs([(u,b),(v,c)]))/k
simplify(F_Z3)

f_Z3 = diff(F_Z3,z)
f_Z3

cond_F_Z3 = (integrate(z*f_Z,(z,a*c,b*c))+integrate(z*f_Z2,(z,b*c,a*d))+integrate(z*f_Z3,(z,a*d,z)))/F_Z3
simplify(cond_F_Z3)

N(cond_F_Z3.subs([(a, .5), (b, 1.5), (c, .4), (d, 1.6), (z,1)]))