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A basket of balls is dropped into a maze. When a ball is dropped into the maze at the top it moves downward, pulled by gravity, through a series of nails. The ball then falls down to a new level where it encounters another nail. Finally, at the bottom of the maze, the ball falls into one of the troughs.

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If enough balls are dropped into the maze, we get a distribution of balls which is a mixture of two normal distributions with parameters: $x_{1}\sim\mathcal{N}(\mu_{1},\sigma_{1})$, $x_{2}\sim\mathcal{N}(\mu_{2},\sigma_{2})$, and $w_{1}=1-w_{2}$ (weight).

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As you can see, there are many troughs. Moreover, there are $\tau$ levels of nails. If $X$ is the number of a trough, we know the probability of our balls falling in a trough lower than $X$ after $\tau$ levels of nails. Let's call the density function $f(\tau)=f^{*}$. Now we can repeat this experiment by adding $t$ levels of nails after the first $\tau$ levels of nails, getting a total of $T=\tau + t$ maze levels. Because nails are randomly distributed, we get a distribution of balls which is a mixture of two normal distributions with different parameters than before. We get the probability density function $f(T)=g^{*}$:

enter image description here

So now we know that, after $\tau$ maze levels, our p.d.f. is $f^{*}$; then we know that, after $T=\tau + t$ maze levels, our p.d.f. is $g^{*}$. My question: what's the p.d.f $f(T - \tau)=f(t)$? In other words, how to obtain the p.d.f. implied by this experiment related to balls falling from level $t$ to level $T$ in this case and under what hypothesis does your answer hold?

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    $\begingroup$ "nails are randomly distributed" — They are? I thought they were arranged in a fixed pattern. What procedure is used to randomly decide where to place the nails? $\endgroup$ – Kodiologist Dec 20 '18 at 14:47
  • $\begingroup$ It seems that nails are arranged so that there will be $n = 15$ random bounces, each left or right with probability about 1/2. Then with the numbers of the bins at the bottom as you have labeled them $(0, 1, \dots, 15).$ we can say that the ball comes to rest at bin $X,$ where $X$ is the number or bounces to the right. So (assuming a little more precision than an actual physical apparatus would have) the implied dist'n is $X \sim \mathsf{Binom}(n=15,\,p=1/2).$ In a classroom demo, one might comment that the dist'n is nearly normal. $\endgroup$ – BruceET Dec 20 '18 at 15:45
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    $\begingroup$ The description could use some history details: Galton's quincunx (circa 1898)(en.wikipedia.org/wiki/Bean_machine). Also, a reference to the claim that the result is a mixture of two normals would be interesting. $\endgroup$ – JimB Dec 20 '18 at 17:30
  • $\begingroup$ @Kodiologist: I'm sorry because the picture may be misleading. Although the picture shows a fixed pattern, it's just an example. Actually playing with the normal mixture parameters can return a normal distribution. So the procedure to place the nails is the one which returns your favourite normal mixture. $\endgroup$ – Lisa Ann Dec 20 '18 at 22:00
  • $\begingroup$ @BruceET: let the levels/steps/random bounces go towards $\infty$. I think this should make the central limit theorem come into play. $\endgroup$ – Lisa Ann Dec 20 '18 at 22:04
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This is an extended comment to simply show my confusion as to the details of the problem.

I think that you need to start with the discrete nature of the problem (at least that's what your figure and initial description implies to me). I appears that there are discrete time steps and just two potential outcomes at each nail.

One way to generalize Galton's device is to assume that at each nail (or vertex) there is a probability of the ball going to the left or right. This would be a Bernoulli random variable. For 5 levels (4 levels of transition and 1 final level) the probabilities might be described as follows:

Galtons device with 5 levels

where $p_{i,j}$ is the conditional probability of the ball going to the left of the nail given that the ball is in position $j$ of row $i$.

If $p_{i,j}=p$, then when there are $n$ rows of nails, the probability of landing in the $k$-th trough is given by $$\binom{n-1}{k-1}(1-p)^{k-1} p^{n-k}$$

When all of the $p_{i,j}$ probabilities are different, then the distribution is a bit more complex and can be multi-modal. But it remains a discrete probability distribution. Here is an example with 5 levels:

Resulting distribution for 5 levels

One might be interested in the distribution of $X_\tau$ (where a ball will land after $\tau$ levels or the joint distribution of $X_\tau$ and $X_{\tau+t}$ but if so, that is not clear (to me) from your description.

You've mentioned continuous probability density functions which might be handy for a large number of levels and varying nail probabilities but that would seem to come into play after the set-up is completely described.

OK. You raise a good question. I'm just not seeing the necessary details to move forward.

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