# Further Still On The Wealth Of Stations

My fellow students and I have spent some time investigating my suspicion that serendipity, beyond worth, might account for the relative fortune of the few over the many. To this end we have set to creating perfectly fair games mimicking the manner in which wealth accumulates amongst the populace, that we might discover whether their outcomes should elevate some small lucky band of players well above their fellows.
Thus far we have seen that games of both random returns and losses of players' funds and random trade between them most certainly do so, but their rules failed to take into account either the value of labour or the cost of sustenance, somewhat weakening any conclusions that we might have drawn from their study.
We have consequently spent some time creating further rules to rectify these deficiencies.

### A Rule Of Labour

In formulating a rule to reflect wealth creation through labour, we postulated that the greater one's assets, the more one might prove able to create wealth from them; the more land one owns, for example, the more food one might produce. Our rule of labour is therefore rather similar to that of our first game in that each player adds to their funds a random proportion of them somewhere between zero and an upper limit. Specifically, each player's funds are updated according to the rule
\begin{align*} u &\sim U(0, b)\\ x &\leftarrow x + u \times x \end{align*}
where $$U(0, b)$$ represents a uniformly distributed random variable with lower and upper bounds of zero and $$b$$ respectively, and $$x$$ that player's funds.
In adding this rule to our trading game we split its play into alternate rounds, the first of which gave every player a turn at labour and the second a turn at trade.

You will no doubt recall that in our trading game each player would in turn contract a number of trades equal to the smallest whole number no less than their funds with randomly chosen players, with each trade exchanging up to some fraction $$c$$ of the lesser of their funds by
\begin{align*} u &\sim U(-c, c)\\ x^\prime_1 &= x_1 - u \times \min\left(x_1, x_2\right)\\ x^\prime_2 &= x_2 + u \times \min\left(x_i, x_2\right) \end{align*}
where $$x_1$$ and $$x_2$$ are the players' funds beforehand and $$x^\prime_1$$ and $$x^\prime_2$$ their funds afterwards.

Deck 1 demonstrates the effect of adding the labour rule, with a return of up to five and one tenth of a percent, to our trading game.

Deck 1: A Game Of Labour And Trade Once again, the players' initial funds are marked by a red line which rapidly falls toward zero, demonstrating that the players typically have significantly more funds at the conclusion of the game than they had at its commencement.
Nevertheless, we still find that a lucky few profit far above their fellows as is emphasised by deck 2 which shows the players' funds from the greatest to the least after each turn.

Deck 2: Sorted By Decreasing Funds To gain some sense of how the outcomes are distributed, we put together deck 3 to construct a histogram of them.

Deck 3: The Histogram Of Outcomes That the players by and large increase their funds many times over in this game is reflected by the fact that the upper bound of the histogram is thousands of millions times larger than their initial funds.
Even so, we can see that it is still far more likely that a player should finish the game amongst the least wealthy than amongst the most.

### A Rule Of Sustenance

In formulating a rule of sustenance, my fellow students and I thought it not unreasonable to set a hard and fast lower limit upon the funds consumed at each turn since one cannot live by breath alone!
To this end, we added a third round to each turn in which each player consumed the sum of a random proportion between zero and $$d$$ of their funds and a fixed quantity $$d_0$$
\begin{align*} u &\sim U(0, d)\\ x &\leftarrow x - \left(u \times x + d_0\right) \end{align*}
Note that, as a consequence of this rule, it is entirely possible that a player's funds might fall below zero, under which circumstances we decided to simply set them to zero.
Deck 4 shows the consequences of our rule of sustenance upon the players' outcomes, with up to five percent of their current funds plus one ten thousandth of their initial funds consumed at each turn.

Deck 4: A Game Of Labour, Trade And Sustenance Whilst the players all unsurprisingly conclude the game a good deal worse off with the addition of this rule, it most certainly seems that it has further elevated the fortunate few above the unfortunate many. To see whether or not this is in fact the case, deck 5 plots a histogram of the players' final funds.

Deck 5: The Histogram Of Outcomes The precipitous drop from the likelihood of the players being amongst the least prosperous to their not being indubitably confirms that the cost of sustenance has had a deleterious effect upon many of their fortunes.

### Reckoning The Players' Chances

Once again, my fellow students and I sought to quantify this observation by figuring the chances that a player might conclude the game with greater than some multiple or less than some fraction of their original funds and to this end we put together deck 6 to play the game many times over and count such outcomes.

Deck 6: The Chances Of Losses And Gains

Evidently the rule of sustenance has led to near ruin for many of their number and great riches for very few of them, despite being applied equally to each and every one!

We also sought to again measure the chances that a player who had fared poorly at the beginning of the game might conclude it with greater funds than one who had fared well by giving halved funds to half of them and doubled funds to the rest, counting how often those amongst the former had funds exceeding those of their counterparts amongst the latter after five hundred turns, as demonstrated by deck 7.

Deck 7: The Chance Of A Comeback

As has consistently been the case with our games, the unfortunate have little chance of turning the tables upon the fortunate by ultimately profiting above them, again highlighting the advantage to be had from a lucky start!

### The Precise Implications Of Sustenance

Naturally, my fellow students and I were keen to discover precisely the effect that these rules might have upon the players' funds at each turn. Alas, we have no more been able to figure the implications of the rule of trade upon this game than we were of it upon our last.
We were, however, able to express the consequences of a single turn of just the rules of labour and sustenance with
\begin{align*} b &\in (0, \infty)\\ d &\in (0, 1)\\ \\ u_1 &\sim U(0, b)\\ u_2 &\sim U(0, d)\\ \\ x_1 &= x_0 \times (1 + u_1)\\ x_2 &= x_1 \times (1 - u_2) - d_0 \end{align*}
where $$x_0$$ represents a player's funds at the start of a turn, $$x_1$$ their funds after the rule of labour and $$x_2$$ those after the rule of sustenance.
To simplify our endeavour, we elected not to figure the outcome of a turn directly, but rather that of
\begin{align*} x_2 &= x_0 \times (1 + u_1) \times (1 - u_2) - d_0\\ z &= \frac{x_2 + d_0}{x_0} = (1 + u_1) \times (1 - u_2) \end{align*}
This we further simplified by choosing to draw from uniform random variables that embodied the addition and subtraction to and from one with
\begin{align*} u_1^\prime &\sim U(1, 1+b)\\ u_2^\prime &\sim U(1-d, 1)\\ z &= u_1^\prime \times u_2^\prime \end{align*}
Now, to figure the outcomes of such products of uniformly distributed random variables we employed logarithms to transform them into sums
\begin{align*} v_1 &= \ln u_1^\prime\\ v_2 &= \ln u_2^\prime\\ \ln z &= \ln \left(u_1^\prime \times u_2^\prime\right) = \ln u_1^\prime + \ln u_2^\prime = v_1 + v_2 \end{align*}
The properties of the logarithm of a uniformly distributed random variable can be deduced from its cumulative distribution function, or CDF
$P_v(x) = \Pr(v \leqslant x) = \Pr(\ln u \leqslant x) = \Pr\left(u \leqslant e^x\right) = P_u\left(e^x\right)$
Specifically, for
\begin{align*} u &\sim U(\alpha, \beta) \quad 0 \leqslant \alpha \leqslant \beta < \infty\\ v &= \ln u\\ \end{align*}
we have
$P_u\left(e^x\right) = \begin{cases} 0 & e^x \in [0, \alpha)\\ \dfrac{e^x - \alpha}{\beta-\alpha} & e^x \in [\alpha, \beta)\\ 1 & e^x \in [\beta, \infty)\\ \end{cases}$
where $$\in$$ means within and $$[\alpha, \beta)$$ represents an interval including the lesser value $$\alpha$$ up to but not including the greater $$\beta$$, and consequently
$P_v(x) = \begin{cases} 0 & x \in (-\infty, \ln \alpha)\\ \dfrac{e^x - \alpha}{\beta-\alpha} & x \in [\ln \alpha, \ln \beta)\\ 1 & x \in [\ln \beta, \infty)\\ \end{cases}$
From this we might find the probability density function, or PDF, by differentiation
$p_v(x) = \begin{cases} 0 & x \in (-\infty, \ln \alpha)\\ \dfrac{e^x}{\beta-\alpha} & x \in [\ln \alpha, \ln \beta)\\ 0 & x \in [\ln \beta, \infty)\\ \end{cases}$
The PDF of the sum of a pair of random variables can be expressed in terms of their PDFs by means of a convolution
$p_{v_1+v_2}(x) = \int_{-\infty}^\infty p_{v_1}(x-t) \times p_{v_2}(t) \; \mathrm{d}t$
For the product being integrated to be not equal to zero requires that both of its terms are not equal to zero, from which we can deduce that
\begin{align*} p_{v_1}(x-t) \neq 0 &\implies x-t \in [\ln 1, \ln (1+b))\\ &\implies \phantom{x}-t \in [\ln 1 - x, \ln (1+b) - x)\\ &\implies \phantom{x-}\,t \in (x - \ln (1+b), x - \ln 1] = (x - \ln (1+b), x] \end{align*}
and
$p_{v_2}(t) \neq 0 \implies t \in [\ln (1-d), \ln 1) = [\ln (1-d), 0)$
where $$\implies$$ means implies.

Putting these conditions together yields
$t \in \begin{cases} [\ln (1-d), x] & x - \ln (1+b) < \ln(1-d) \wedge x < 0 \wedge x \geqslant \ln(1-d)\\ [\ln (1-d), 0) & x - \ln (1+b) < \ln(1-d) \wedge x \geqslant 0\\ (x - \ln (1+b), x] & x - \ln (1+b) \geqslant \ln(1-d) \wedge x < 0\\ (x - \ln (1+b), 0) & x - \ln (1+b) \geqslant \ln(1-d) \wedge x \geqslant 0 \wedge x - \ln (1+b) < 0\\ \varnothing & \mathrm{otherwise} \end{cases}$
where $$\wedge$$ means and and $$\varnothing$$ represents an empty interval, which upon some small rearrangement further yields
$t \in \begin{cases} [\ln (1-d), x] & x < \ln(1-d) + \ln (1+b) \wedge x < 0 \wedge x \geqslant \ln(1-d)\\ [\ln (1-d), 0) & x < \ln(1-d) + \ln (1+b) \wedge x \geqslant 0\\ (x - \ln (1+b), x] & x \geqslant \ln(1-d) + \ln (1+b) \wedge x < 0\\ (x - \ln (1+b), 0) & x \geqslant \ln(1-d) + \ln (1+b) \wedge x \geqslant 0 \wedge x < \ln (1+b)\\ \varnothing & \mathrm{otherwise} \end{cases}$
Now, if $$\ln(1-d) + \ln (1+b)$$ is less than zero then it is trivially the case that
$x < \ln(1-d) + \ln (1+b) \implies x < 0$
and so, in such circumstances, we may simplify the extent of $$t$$ for which the product is non-zero to
$t \in \begin{cases} [\ln (1-d), x] & x < \ln(1-d) + \ln (1+b) \wedge x \geqslant \ln(1-d)\\ (x - \ln (1+b), x] & x \geqslant \ln(1-d) + \ln (1+b) \wedge x < 0\\ (x - \ln (1+b), 0) & x \geqslant 0 \wedge x < \ln (1+b)\\ \varnothing & \mathrm{otherwise} \end{cases}$
Similarly, if $$\ln(1-d) + \ln (1+b)$$ is greater than or equal to zero then
$x \geqslant \ln(1-d) + \ln (1+b) \implies x \geqslant 0$
yielding
$t \in \begin{cases} [\ln (1-d), x] & x < 0 \wedge x \geqslant \ln(1-d)\\ [\ln (1-d), 0) & x < \ln(1-d) + \ln (1+b) \wedge x \geqslant 0\\ (x - \ln (1+b), 0) & x \geqslant \ln(1-d) + \ln (1+b) \wedge x < \ln (1+b)\\ \varnothing & \mathrm{otherwise} \end{cases}$
Consequently, we have
\begin{align*} \ln(1-d) + \ln (1+b) < 0 &\implies t \in \begin{cases} [\ln (1-d), x] & x \in [\ln(1-d), \ln(1-d) + \ln (1+b))\\ (x - \ln (1+b), x] & x \in [\ln(1-d) + \ln (1+b), 0)\\ (x - \ln (1+b), 0) & x \in [0, \ln (1+b))\\ \varnothing & \mathrm{otherwise} \end{cases}\\ \\ \ln(1-d) + \ln (1+b) \geqslant 0 &\implies t \in \begin{cases} [\ln (1-d), x] & x \in [\ln(1-d), 0)\\ [\ln (1-d), 0) & x \in [0, \ln(1-d) + \ln (1+b))\\ (x - \ln (1+b), 0) & x \in [\ln(1-d) + \ln (1+b), \ln (1+b))\\ \varnothing & \mathrm{otherwise} \end{cases} \end{align*}
Furthermore, within a particular interval from $$t_1$$ to $$t_2$$, the convolution reckons to
\begin{align*} \int_{t_1}^{t_2} p_{v_1}(x-t) \times p_{v_2}(t) \; \mathrm{d}t &= \int_{t_1}^{t_2} \frac{e^{x-t}}{(b+1) - 1} \times \frac{e^t}{1 - (1-d)} \; \mathrm{d}t\\ &= \int_{t_1}^{t_2} \frac{e^x}{b \times d} \mathrm{d}t\\ &= \left[\frac{e^x \times t}{b \times d}\right]_{t_1}^{t_2}\\ &= \frac{e^x}{b \times d} \times \left(t_2 - t_1\right) \end{align*}

### Case The First

For negative $$\ln(1-d) + \ln(1+b)$$ this yields the PDF
\begin{align*} p_{v_1 + v_2}(x) &= \begin{cases} \frac{e^x}{b \times d} \times \left(x - \ln (1-d)\right) & x \in [\ln(1-d), \ln(1-d) + \ln (1+b))\\ \frac{e^x}{b \times d} \times \left(x - \left(x - \ln (1+b)\right)\right) & x \in [\ln(1-d) + \ln (1+b), 0)\\ \frac{e^x}{b \times d} \times \left(0 - \left(x - \ln (1+b)\right)\right) & x \in [0, \ln (1+b))\\ 0 & \mathrm{otherwise} \end{cases}\\ \\ &= \begin{cases} \frac{e^x}{b \times d} \times \left(x - \ln (1-d)\right) & x \in [\ln(1-d), \ln(1-d) + \ln (1+b))\\ \frac{e^x}{b \times d} \times \ln (1+b) & x \in [\ln(1-d) + \ln (1+b), 0)\\ \frac{e^x}{b \times d} \times \left(\ln (1+b) - x\right) & x \in [0, \ln (1+b))\\ 0 & \mathrm{otherwise} \end{cases} \end{align*}
To recover the CDF from this PDF we must integrate each of its cases within their intervals of applicability. For example, the integral of the first is given by
$f_1(x) = \int_{\ln(1-d)}^x \frac{e^t}{b \times d} \times \left(t - \ln (1-d)\right) \; \mathrm{d}t$
To resolve this integral we must use the rule of integration by parts which states that
$\int \frac{\mathrm{d}}{\mathrm{d}t} g(t) \times h(t) \; \mathrm{d}t = \bigg[g(t) \times h(t)\bigg] - \int g(t) \times \frac{\mathrm{d}}{\mathrm{d}t} h(t) \; \mathrm{d}t$
In particular, if we choose
\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} g(t) &= \frac{e^t}{b \times d}\\ h(t) &= t - \ln(1-d) \end{align*}
then we have
\begin{align*} g(t) &= \frac{e^t}{b \times d}\\ \frac{\mathrm{d}}{\mathrm{d}t} h(t) &= 1 \end{align*}
and consequently
\begin{align*} f_1(x) &= \bigg[\frac{e^t}{b \times d} \times \left(t - \ln (1-d)\right)\bigg]_{\ln(1-d)}^x - \int_{\ln(1-d)}^x \frac{e^t}{b \times d} \times 1 \; \mathrm{d}t\\ &= \bigg[\frac{e^t}{b \times d} \times \left(t - \ln (1-d)\right)\bigg]_{\ln(1-d)}^x - \bigg[\frac{e^t}{b \times d}\bigg]_{\ln(1-d)}^x\\ &= \bigg[\frac{e^t}{b \times d} \times \left(t - \ln (1-d) - 1\right)\bigg]_{\ln(1-d)}^x\\ &= \left(\frac{e^x}{b \times d} \times \left(x - \ln (1-d) - 1\right)\right) - \left(\frac{e^{\ln(1-d)}}{b \times d} \times \left(\ln(1-d) - \ln (1-d) - 1\right)\right)\\ &= \frac{e^x}{b \times d} \times \left(x - \ln (1-d) - 1\right) - \frac{1-d}{b \times d} \times -1\\ &= \frac{e^x}{b \times d} \times \left(x - \ln (1-d) - 1\right) + \frac{1-d}{b \times d} \end{align*}
Similarly, for the second case we have
$f_2(x) = \int_{\ln(1-d) + \ln (1+b)}^x \frac{e^t}{b \times d} \times \ln (1+b) \; \mathrm{d}t = \frac{e^x}{b \times d} \times \ln (1+b) - \frac{(1-d) \times (1+b)}{b \times d} \times \ln (1+b)$
and for the third
$f_3(x) = \int_0^x \frac{e^t}{b \times d} \times \left(\ln (1+b) - t\right) \; \mathrm{d}t = \frac{e^x}{b \times d} \times \left(\ln (1+b) - x + 1\right) - \frac{1}{b \times d} \times \left(\ln (1+b) + 1\right)$
In defining the CDF we must take care to include the integrals over every interval that falls entirely beneath any particular point of interest. To this end, if we define the constants
\begin{align*} c_1 &= f_1\left(\ln(1-d) + \ln(1+b)\right)\\ c_2 &= f_2\left(0\right)\\ c_3 &= f_3\left(\ln(1+b)\right) \end{align*}
then we may express the CDF as
$P_{v_1 + v_2}(x) = \begin{cases} 0 & x \in (-\infty, \ln(1-d))\\ f_1(x) & x \in [\ln(1-d), \ln(1-d) + \ln (1+b))\\ c_1 + f_2(x) & x \in [\ln(1-d) + \ln (1+b), 0)\\ c_1 + c_2 + f_3(x) & x \in [0, \ln (1+b))\\ c_1 + c_2 + c_3 & x \in [\ln (1+b), \infty) \end{cases}$
Expanding out $$c_1$$ yields
\begin{align*} c_1 &= \frac{e^{\ln(1-d) + \ln(1+b)}}{b \times d} \times \left(\ln(1-d) + \ln(1+b) - \ln (1-d) - 1\right) + \frac{1-d}{b \times d}\\ &= \frac{(1-d) \times (1+b)}{b \times d} \times \left(\ln(1+b) - 1\right) + \frac{1-d}{b \times d}\\ &= \frac{(1-d) \times (1+b)}{b \times d} \times \ln(1+b) - \frac{(1-d) \times (1+b)}{b \times d} + \frac{1-d}{b \times d}\\ &= \frac{(1-d) \times (1+b)}{b \times d} \times \ln(1+b) - \frac{(1-d) \times (1+b) - (1-d)}{b \times d}\\ &= \frac{(1-d) \times (1+b)}{b \times d} \times \ln(1+b) - \frac{(1-d) \times b}{b \times d}\\ &= \frac{(1-d) \times (1+b)}{b \times d} \times \ln(1+b) - \frac{1-d}{d}\\ &= \frac{(1-d) \times (1+b)}{b \times d} \times \ln(1+b) + 1 - \frac{1}{d} \end{align*}
and $$c_2$$
\begin{align*} c_2 &= \frac{e^0}{b \times d} \times \ln (1+b) - \frac{(1-d) \times (1+b)}{b \times d} \times \ln (1+b)\\ &= \frac{1}{b \times d} \times \ln (1+b) - \frac{(1-d) \times (1+b)}{b \times d} \times \ln (1+b) \end{align*}
and, finally, $$c_3$$
\begin{align*} c_3 &= \frac{e^{\ln(1+b)}}{b \times d} \times \left(\ln (1+b) - \ln (1+b) + 1\right) - \frac{1}{b \times d} \times \left(\ln (1+b) + 1\right)\\ &= \frac{1+b}{b \times d} \times 1 - \frac{1}{b \times d} \times \left(\ln (1+b) + 1\right)\\ &= \frac{1+b}{b \times d} - \frac{1}{b \times d} - \frac{1}{b \times d} \times \ln (1+b)\\ &= \frac{1}{d} - \frac{1}{b \times d} \times \ln (1+b) \end{align*}
Now, the sum of $$c_1$$ and $$c_2$$ is trivially equal to
$c_1 + c_2 = \frac{1}{b \times d} \times \ln (1+b) + 1 - \frac{1}{d}$
and that of $$c_1$$, $$c_2$$ and $$c_3$$ to
$c_1 + c_2 + c_3 = 1$
which I must say came as something of a relief upon our finally having figured it, since the CDF must take a greatest value of one!

Finally, the third and fourth cases of the CDF resolve to
$c_1 + f_2(x) = \frac{e^x}{b \times d} \times \ln (1+b) + 1 - \frac{1}{d}$
and
$c_1 + c_2 + f_3(x) = \frac{e^x}{b \times d} \times \left(\ln (1+b) - x + 1\right) + 1 - \frac{1+b}{b \times d}$
yielding the whole
$P_{v_1 + v_2}(x) = \begin{cases} 0 & x \in (-\infty, \ln(1-d))\\ \frac{e^x}{b \times d} \times \left(x - \ln (1-d) - 1\right) + \frac{1-d}{b \times d} & x \in [\ln(1-d), \ln(1-d) + \ln (1+b))\\ \frac{e^x}{b \times d} \times \ln (1+b) + 1 - \frac{1}{d} & x \in [\ln(1-d) + \ln (1+b), 0)\\ \frac{e^x}{b \times d} \times \left(\ln (1+b) - x + 1\right) + 1 - \frac{1+b}{b \times d} & x \in [0, \ln (1+b))\\ 1 & x \in [\ln (1+b), \infty) \end{cases}$
To deduce from this the CDF of our transformed outcome $$z$$ we need simply exploit the properties of the CDF once again with
\begin{align*} \ln z &= v_1 + v_2\\ P_{z}(x) &= \Pr\left(z \leqslant x\right) = \Pr\left(\ln z \leqslant \ln x\right) = \Pr\left(v_1 + v_2 \leqslant \ln x\right) = P_{v_1 + v_2}(\ln x) \end{align*}
and we therefore have
\begin{align*} P_{z}(x) &= \begin{cases} 0 & \ln x \in (-\infty, \ln(1-d))\\ \frac{e^{\ln x}}{b \times d} \times \left(\ln x - \ln (1-d) - 1\right) + \frac{1-d}{b \times d} & \ln x \in [\ln(1-d), \ln(1-d) + \ln (1+b))\\ \frac{e^{\ln x}}{b \times d} \times \ln (1+b) + 1 - \frac{1}{d} & \ln x \in [\ln(1-d) + \ln (1+b), 0)\\ \frac{e^{\ln x}}{b \times d} \times \left(\ln (1+b) - \ln x + 1\right) + 1 - \frac{1+b}{b \times d} & \ln x \in [0, \ln (1+b))\\ 1 & \ln x \in [\ln (1+b), \infty) \end{cases}\\ \\ &= \begin{cases} 0 & x \in (-\infty, 1-d)\\ \frac{x}{b \times d} \times \left(\ln x - \ln (1-d) - 1\right) + \frac{1-d}{b \times d} & x \in [1-d, (1-d) \times (1+b))\\ \frac{x}{b \times d} \times \ln (1+b) + 1 - \frac{1}{d} & x \in [(1-d) \times (1+b), 1)\\ \frac{x}{b \times d} \times \left(\ln (1+b) - \ln x + 1\right) + 1 - \frac{1+b}{b \times d} & x \in [1, 1+b)\\ 1 & x \in [1+b, \infty) \end{cases} \end{align*}
As before we must differentiate these cases to recover the PDF of $$z$$, for which we must employ the product rule, which states that
$\frac{\mathrm{d}}{\mathrm{d}x} \left(f(x) \times g(x)\right) = \left(\frac{\mathrm{d}}{\mathrm{d}x} f(x)\right) \times g(x) + f(x) \times \left(\frac{\mathrm{d}}{\mathrm{d}x} g(x)\right)$
For example, if for the second case we choose
\begin{align*} f(x) &= \frac{x}{b \times d}\\ g(x) &= \ln x - \ln (1-d) - 1 \end{align*}
then we find that
\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} f(x) &= \frac{1}{b \times d}\\ \frac{\mathrm{d}}{\mathrm{d}x} g(x) &= \frac{1}{x} \end{align*}
and so, noting that the derivative of any constant term must equal zero, its derivative is given by
\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left(f(x) \times g(x)\right) &= \frac{1}{b \times d} \times \left(\ln x - \ln (1-d) - 1\right) + \frac{x}{b \times d} \times \frac{1}{x}\\ &= \frac{1}{b \times d} \times \left(\ln x - \ln (1-d) - 1\right) + \frac{1}{b \times d}\\ &= \frac{1}{b \times d} \times \left(\ln x - \ln (1-d)\right) \end{align*}
and the PDF is consequently given by
$p_{z}(x) = \begin{cases} 0 & x \in (-\infty, 1-d)\\ \frac{1}{b \times d} \times \left(\ln x - \ln (1-d)\right) & x \in [1-d, (1-d) \times (1+b))\\ \frac{1}{b \times d} \times \ln (1+b) & x \in [(1-d) \times (1+b), 1)\\ \frac{1}{b \times d} \times \left(\ln (1+b) - \ln x\right) & x \in [1, 1+b)\\ 0 & x \in [1+b, \infty) \end{cases}$

### Case The Second

The reckoning of the governing distribution of $$z$$ when $$\ln(1-d) + \ln(1+b)$$ is greater than or equal to zero proceeds in much the same rather tedious fashion, ultimately yielding the CDF
$P_z(x) = \begin{cases} 0 & x \in [0, 1-d)\\ \frac{x}{b \times d} \times (\ln x - \ln(1-d) - 1) + \frac{1-d}{b \times d} & x \in [1-d, 1)\\ -\frac{x}{b \times d} \times \ln(1-d) - \frac{1}{b} & x \in [1, (1-d) \times (1+b))\\ \frac{x}{b \times d} \times (\ln(1+b) - \ln x + 1) + 1 - \frac{1+b}{b \times d} & x \in [(1-d) \times (1+b), 1+b)\\ 1 & x \in [1+b, \infty) \end{cases}$
and the PDF
$p_z(x) = \begin{cases} 0 & x \in [0, 1-d)\\ \frac{1}{b \times d} \times (\ln x - \ln(1-d)) & x \in [1-d, 1)\\ -\frac{1}{b \times d} \times \ln(1-d) & x \in [1, (1-d) \times (1+b))\\ \frac{1}{b \times d} \times (\ln(1+b) - \ln x) & x \in [(1-d) \times (1+b), 1+b)\\ 0 & x \in [1+b, \infty) \end{cases}$

### The Distribution Of Untransformed Outcomes

Given these results, we can easily recover the distribution of the untransformed outcomes by exploiting the fact that
\begin{align*} z &= \frac{x_2 + d_0}{x_0}\\ P_{x_2}(x) &= \Pr\left(x_2 \leqslant x\right) = \Pr\left(z \leqslant \frac{x + d_0}{x_0}\right) = P_z\left(\frac{x + d_0}{x_0}\right)\\ p_{x_2}(x) &= \frac{\mathrm{d}}{\mathrm{d}x} P_{x_2}(x) = \frac{\mathrm{d}}{\mathrm{d}x} P_z\left(\frac{x + d_0}{x_0}\right) = \frac{1}{x_0} p_z\left(\frac{x + d_0}{x_0}\right) \end{align*}
Noting that
$\ln(1-d) + \ln(1+b) < 0 \implies (1-d) \times (1+b) < 1$
and
$\ln(1-d) + \ln(1+b) \geqslant 0 \implies (1-d) \times (1+b) \geqslant 1$
my fellow students and I formulated script 1 to figure values of the CDF

Script 1: The CDF Of Outcomes
function cdf(x, x0, b, d, d0) {
var m = (1-d)*(1+b);
var z = (x+d0)/x0;
var c;

if(m<1) {
if(z<1-d)      c = 0;
else if(z<m)   c = z*(Math.log(z)-Math.log(1-d)-1)/(b*d) + (1-d)/(b*d);
else if(z<1)   c = z*Math.log(1+b)/(b*d) + 1-1/d;
else if(z<1+b) c = z*(Math.log(1+b)-Math.log(z)+1)/(b*d) + 1 - (1+b)/(b*d);
else           c = 1;
}
else {
if(z<1-d)      c = 0;
else if(z<1)   c = z*(Math.log(z)-Math.log(1-d)-1)/(b*d) + (1-d)/(b*d);
else if(z<m)   c = -z*Math.log(1-d)/(b*d) - 1/b;
else if(z<1+b) c = z*(Math.log(1+b)-Math.log(z)+1)/(b*d) + 1 - (1+b)/(b*d);
else           c = 1;
}
return c;
}


and script 2 to figure those of the PDF

Script 2: The PDF Of Outcomes
function pdf(x, x0, b, d, d0) {
var m = (1-d)*(1+b);
var z = (x+d0)/x0;
var p;

if(m<1) {
if(z<1-d)      p = 0;
else if(z<m)   p = (Math.log(z)-Math.log(1-d))/(b*d);
else if(z<1)   p = Math.log(1+b)/(b*d);
else if(z<1+b) p = (Math.log(1+b)-Math.log(z))/(b*d);
else           p = 0;
}
else {
if(z<1-d)      p = 0;
else if(z<1)   p = (Math.log(z)-Math.log(1-d))/(b*d);
else if(z<m)   p = -Math.log(1-d)/(b*d);
else if(z<1+b) p = (Math.log(1+b)-Math.log(z))/(b*d);
else           p = 0;
}
return p/x0;
}


To satisfy ourselves that we had correctly reckoned the consequences of the rules of labour and sustenance we put together deck 8 to compare a histogram of their outcomes with a graph of their likelihoods as predicted by our CDF.

Deck 8: Verifying The CDF That they align so closely we took as compelling evidence that our ratiocination had been sound!

The CDF sheds some light upon the reason why many of the players fared so poorly in the game; the probability that a player should fail to profit at the conclusion of a turn is significantly greater if they began it with relatively few funds, as demonstrated by deck 9 which plots how the probability of loss decreases as funds increase.

Deck 9: The Probability Of Loss This is further illuminated by considering the relative return that a player should expect after each turn
$\mathrm{E}\left[\frac{x_2 - x_0}{x_0}\right]$
We can express this in terms of the expected value of $$z$$ by noting that
$\mathrm{E}(z) = \mathrm{E}\left[\frac{x_2 + d_0}{x_0}\right] = \mathrm{E}\left[\frac{x_2}{x_0}\right] + \frac{d_0}{x_0} = \mathrm{E}\left[\frac{x_2 - x_0}{x_0}\right] + 1 + \frac{d_0}{x_0}$
and so
$\mathrm{E}\left[\frac{x_2 - x_0}{x_0}\right] = \mathrm{E}(z) - 1 - \frac{d_0}{x_0}$
The PDF of $$z$$ does not depend upon $$x_0$$ and so, consequently, nor does its expected value. The only term in the expected return that does is therefore the last, which grows negative without limit as $$x_0$$ approaches zero; a most unfortunate prospect for any already impoverished players!

### In Conclusion

Once again we have seen that a game of perfectly equitable rules favours the lucky few well above the unfortunate many. The question that my fellow students and I should finally like answered is whether it is possible to introduce a rule that mitigates the capricious hand of providence and, when our studies permit, we shall be sure to address it!
$$\Box$$

### References

 On The Wealth Of Stations, www.thusspakeak.com, 2016.

 Further On The Wealth Of Stations, www.thusspakeak.com, 2016.

### Gallimaufry  AKCalc
ECMA  Endarkenment
Turning Sixteen

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