{"id":4116,"date":"2022-09-14T12:26:42","date_gmt":"2022-09-14T10:26:42","guid":{"rendered":"https:\/\/erlebnisland-mathematik.de\/?page_id=4116"},"modified":"2023-11-10T13:21:07","modified_gmt":"2023-11-10T12:21:07","slug":"advanced-text-money-clay-board","status":"publish","type":"page","link":"https:\/\/erlebnisland-mathematik.de\/en\/advanced-text-money-clay-board\/","title":{"rendered":"Advanced Text Money Clay Board"},"content":{"rendered":"

[vc_row drowwidth=”sidebar-biest-default sidebar-biest”][vc_column][vc_column_text]<\/p>\n

Galton Board<\/h1>\n

At the end of the 19th century, the English polymath Sir Francis C. Galton<\/em> (1822–1911) developed an arrangement to demonstrate the so-called binomial distribution.<\/em> This arrangement was later called the Galton<\/em> board in his honour. A version of this has been realised in ADVENTURE LAND MATHEMATICS:<\/p>\n

Between two glass plates, several 50-cent coins are fixed with three pins each and arranged evenly so that — as an overall structure — an equilateral triangle of twelve “cascades<\/em>” results (see figure 1 below):[\/vc_column_text][vc_single_image image=”1507″ img_size=”large” alignment=”center”]Figure 1: Schematic representation of the Galton board[\/vc_single_image][vc_column_text]If you now let a coin fall in vertically from above, each of these obstacles (i.e. the locked 50-cent coins) randomly<\/em> decides whether it falls to the right or to the left. The probability<\/em> of this is \"p=0.5\" in each case. Below the Galton board are several compartments in which the inserted coins are stacked on top of each other. That is, in this way a kind of bar chart<\/em> is created that approximates the density of a normal distribution<\/em> as the number of inserted coins increases. The shape of this density is a bell-shaped curve (“Gaussian bell curve<\/em>“), as shown in the following figure 2:[\/vc_column_text][vc_single_image image=”1511″ img_size=”large” alignment=”center”]Figure 2: Gaussian bell curve[\/vc_single_image][vc_column_text]1-cent, 2-cent, … or 50-cent coins are suitable for carrying out the experiment in ADVENTURE LAND MATHEMATICS.<\/p>\n

Note<\/em>: The coins thrown in for an experiment on the Galton board will later be used as a contribution to the work of the non-profit(!) association for the promotion of the work of ADVENTURE LAND MATHEMATICS.<\/em> This is how the name “Geldtonbrett<\/em>” came about for this exhibit.[\/vc_column_text][vc_column_text]<\/p>\n

And now … the mathematics:<\/h3>\n

[\/vc_column_text][vc_column_text]The Galton board in ADVENTURE LAND MATHEMATICS has a total of \"N=12\" steps, i.e. the inserted coin has to “choose<\/em>” right or left exactly twelve times on its way down. It is reasonable to assume that all these decisions are stochastically independent<\/em> of each other: Whether the coin decides one way or the other at an obstacle — it does not affect its behaviour below. So essentially the same random experiment<\/em> is run twelve times in a row. So accordingly there is exactly <\/p>\n

  <\/span>   <\/span>\"\[2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2=2^{12}d= 4.096\]\"<\/p>\n

Ways in which the coin can behave. Each of these possibilities corresponds to exactly one path in the associated tree diagram<\/em>. However, the experiment does not count how often a ball uses each of these (equally probable) paths, but how often it chooses right and left.<\/p>\n

For this purpose, below the Galton board \"N+1=13\" are compartments \"F_0, F_1,\dots, F_{12}\" reflecting this. The coin lands in the compartment \"F_n\" exactly when it decides exactly \"n\" times for the right (and accordingly \"(N-n)=(13-n)\" times for the left).<\/p>\n

If one now conducts this experiment with a sufficiently large number of coins, it can be observed that most of them collect in the middle compartments. The outside compartments, on the other hand, are rarely reached.<\/p>\n

This in turn has a simple mathematical reason: The number of paths that lead into the compartment \"F_n\" is exactly \"{N \choose n}={12 \choose n}\", because it is necessary to select exactly \"n\" of the \"N=12\" steps where the coin decides to go right. Accordingly, the probability of ending up in compartment \"F_n\" is exactly <\/p>\n

  <\/span>   <\/span>\"\[p_N(n) ={N \choose n}\cdot p^n\cdot(1-p)^{N-n}={N \choose n}2^{-N}={N \choose n}2^{-12}\]\"<\/p>\n

where \"n=0,\ldots,12\".<\/p>\n

This probability distribution is called a binomial distribution<\/em> with the parameters \"p= 0.5\" (the probability of a coin falling to the right at an obstacle) and \"N=12\" (the number of stages of the random experiment). The \"p_N(n)\" are also called their individual probabilities<\/em>.<\/p>\n

The fact that the binomial distribution for “large<\/em>” values \"N\" is approximated by the normal distribution<\/em> — i.e. is approximated<\/em> — i.e. approximated — by the normal distribution, describes the so-called limit theorem of Moivre-Laplace<\/em>: <\/p>\n

  <\/span>   <\/span>\"\[\sum_{n:a\leq\frac{n-\mu}{\sigma}\leq b}{p_N(n)}\to\frac{1}{\sqrt{2\pi}}\int_a^b{e^{-z^2/2}}dz\]\"<\/p>\n

for \"N\to\infty\", \"a,b\in\mathbb R\" fixed and arbitrary, \"\mu=Np\" and \"\sigma=\sqrt{Np(1-p)}\".<\/p>\n

In the following, we sketch another proof of this theorem:<\/p>\n

First, one observes that the density function <\/p>\n

  <\/span>   <\/span>\"\[z\mapsto f(z)=\frac{e^{-z^2/2}}{\sqrt{2\pi}}\]\"<\/p>\n

has the derivative \"-zf(z)\", i.e. satisfies the differential equation \"f'(z)=-z f(z)\". If we add that the integral <\/p>\n

  <\/span>   <\/span>\"\[\int_{-\infty}^\infty f(z)dz=1\]\"<\/p>\n

then the density function of the standard normal distribution\u00a0<\/p>\n

  <\/span>   <\/span>\"\[f(z)=\frac{e^{-z^2/2}}{\sqrt{2\pi}}\]\"<\/p>\n

is obtained\u00a0as a unique solution to this differential equation. The assertion of Moivre-Laplace’s limit theorem is now that the (suitably rescaled) “density function<\/em>” of the binomial distribution with parameters \"p\" and \"N\" converges against this function \"f(z)\". By “rescale<\/em>” we mean here that the \"x\" coordinate is first shifted to the left by \"\mu=pN\" and then the \"x\" axis is compressed by \"\sigma=\sqrt{Np(1-p)}\" and the \"y\" axis is stretched by \"\sigma\". We thus obtain from the binomial distributed random variable \"B_{N,p}\" with values in \"\{0,\ldots,N\}\", the random variable \"B'_{N,p}\" which has values in <\/p>\n

  <\/span>   <\/span>\"\[\left\{\frac{-Np}{\sqrt{Np(1-p)}},\ldots,\frac{N(1-p)}{\sqrt{Np(1-p)}}\right\}.\]\"<\/p>\n

This is a discrete random variable with distribution <\/p>\n

  <\/span>   <\/span>\"\[\sum_{k=0}^N p_N(k)\delta_{\frac{k-\mu}{\sigma}},\]\"<\/p>\n

where \"\delta_x\" is the Dirac measure<\/em><\/a> in \"x\in\mathbb R\". Since the random variable \"B'_{N,p}\" is discrete, it has no density function in the true sense. But we can replace the Dirac measure \"\delta_x\" in the formula just noted by a measure of mass one \"\lambda_x=\sigma\lambda\rvert_{[x-1/(2\sigma),x+1/(2\sigma)]}\" (where \"\lambda\" denotes the Lebesgue measure)<\/em><\/a> equally distributed on \"[x-\sigma/2,x+\sigma/2]\". In this way we obtain another random variable \"B''_{N,p}\", which has a density function. It is easy to see that the distributions of \"B'_{N,p}\" and \"B''_{N,p}\" are weakly equivalent in the limit, i.e. their difference converges weakly<\/a> to \"0\" for \"N\to\infty\". So we can work with \"B''_{N,p}\" instead of \"B'_{N,p}\". To show that their distribution function weakly tends to \"f\lambda\" (\"\lambda\" the Lebesgue measure), we now simply show that it satisfies the above differential equation \"f'(z)=-zf(z)\" in the limit. So let us write \"f_0\" for the density function of \"B''_{N,p}\" and assume that \"z=\frac{k-\mu}{\sigma}\" for a \"k\in\{0,\ldots,N\}\". This certainly makes sense, since such values fill the whole axis with increasing \"N\" and lie ever closer together. Then we have <\/p>\n

  <\/span>   <\/span>\"\[-z=-\frac{k-\mu}{\sigma}\]\"<\/p>\n

and <\/p>\n

  <\/span>   <\/span>\"\[f_0(z)=\sigma p_N(k).\]\"<\/p>\n

But what is \"f_0'(z)\". In fact, \"f_0\" is constant in an environment of \"z\", so \"f_0'(z)=0\" should actually apply. But at the distance \"1/(2\sigma)\" from \"z\" \"f_0\" suddenly makes a jump; so there \"\lvert f_0'(z)\rvert=\infty\" would be. To account for both, we use a difference quotient<\/em> instead of a differential quotient<\/em>: <\/p>\n

  <\/span>   <\/span>\"\[f_0'(z)\approx\frac{f_0(z+1/\sigma)-f_0(z)}{1/\sigma}=\frac{\sigma p_N(k+1)-\sigma p_N(k)}{1/\sigma}=\sigma^2(p_N(k+1)-p_N(k)).\quad(\ast)\]\"<\/p>\n

Thus, for the left-hand side of the differential equation <\/p>\n

  <\/span>   <\/span>\"\[f_0'(z)=\sigma^2(p_N(k+1)-p_N(k))\]\"<\/p>\n

we obtain<\/p>\n

and for the right-hand side <\/p>\n

  <\/span>   <\/span>\"\[-zf_0(z)=-\frac{k-\mu}{\sigma}\sigma p_N(k)=-(k-\mu)p_N(k)\quad(\ast).\]\"<\/p>\n

So, to check the above differential equation in the limit, it suffices — taking into account \"(\ast)\" and \"(\ast\ast)\" — to prove the equation <\/p>\n

  <\/span>   <\/span>\"\[-\frac{f_0'(z)}{zf_0(z)}=-\frac{p_N(k+1)-p_N(k)}{p_N(k)}\cdot\frac{\sigma^2}{k-\mu}\to 1\]\"<\/p>\n

for \"N\to\infty\". We leave this calculation to the reader and refer to the proof on Wikipedia<\/a> where it is carried out. Since also \"\int_{-\infty}^\infty f_0(z)dz=1\", we are done. Of course, this sketch is not a rigorous proof — but it can be transformed into one with some effort.[\/vc_column_text][vc_column_text]<\/p>\n

Literature<\/h3>\n

[\/vc_column_text][vc_column_text][1] Henze, N.: Stochastik f\u00fcr Einsteiger. Eine Einf\u00fchrung in die faszinierende Welt des Zufalls,<\/cite> Springer Spektrum, 10. Auflage, Wiesbaden, 2013.<\/span><\/p>\n

[2] https:\/\/en.wikipedia.org\/wiki\/De_Moivre%E2%80%93Laplace_theorem<\/a>.[\/vc_column_text][\/vc_column][\/vc_row]<\/p>\n<\/div>","protected":false},"excerpt":{"rendered":"

[vc_row drowwidth=”sidebar-biest-default sidebar-biest”][vc_column][vc_column_text] Galton Board At the end of the 19th century, the English polymath Sir Francis C. Galton (1822–1911) developed an arrangement to demonstrate the so-called binomial distribution. This arrangement was later called the Galton board in his honour. A version of this has been realised in ADVENTURE LAND MATHEMATICS: Between two glass plates, … Advanced Text Money Clay Board<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_acf_changed":false,"footnotes":""},"folder":[],"class_list":["post-4116","page","type-page","status-publish","hentry"],"acf":[],"_links":{"self":[{"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/pages\/4116","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/comments?post=4116"}],"version-history":[{"count":14,"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/pages\/4116\/revisions"}],"predecessor-version":[{"id":4810,"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/pages\/4116\/revisions\/4810"}],"wp:attachment":[{"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/media?parent=4116"}],"wp:term":[{"taxonomy":"folder","embeddable":true,"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/folder?post=4116"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}