% one figure, bob edit 9-jun-2005 incomplete \documentclass{ws-procs975x65} %\documentclass{article} \setcounter{footnote}{0} \renewcommand\thefootnote{\arabic{footnote}} \def\addressfont{\normalsize\rmfamily{}} \def\fermisection#1{\bigskip \S\,\,#1.} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \title{\large\bf 38b.\\ A THEOREM OF CALCULATION OF PROBABILITY\\ AND SOME OF ITS APPLICATIONS } \date{} \address{ Teacher's Diploma Thesis of the Scuola Normale di Pisa.\\ Presented on June 20, 1922.\$6pt] Un teorema di calcolo delle probabilit\`a ed alcune sue applicazioni"\\ Tesi di Abilitazione della Scuola Normale Superiore di Pisa } \maketitle \typeout{Italics should not be present in lines after title!} \fermisection{1} The theorem we want to deal with concerns the properties of the sums of many incoherent addenda having a known statistical distribution. The fundamental theorem on these sums is due to Laplace\footnote{Th\'eorie analytique des probabilit\'es, Oeuvres, VII, p.~309.}. We state the theorem together with a short account of its demonstration from which we shall move to establish a new theorem. Let \textit{n} be a very great number and let y_{1},y_{2}\ldots y_{n} represent \textit{n} unknowns, for each of which we know the statistical distribution; that is, we know that the probability that y_{i} has a value ranging between y_{i} and y_{i}+dy_{i} is \varphi_{i}(y_{i})dy_{i}, where \varphi_{i} is a known function for which, obviously \begin{equation} \int^{\infty}_{-\infty}\varphi_{i}(y)dy=1\ , \label{eq_1} \end{equation} which means that y_{i} must certainly have a value between -\infty and +\infty. In addition we will assume that the statistical distribution of y_{i} is not affected by the values that the other \textit{y}'s can assume, that is, we assume the y_{i}'s are completely incoherent among themselves. Then we take the average of the y_{i} to be zero, namely: \begin{equation} \bar{y}_{i}=\int^{\infty}_{-\infty}y\varphi_{i}(y)dy=0\ . \label{eq_2} \end{equation} Finally we set the average of the squared y_{i} to be \begin{equation} \bar{y}^{2}_{i}=\int^{\infty}_{-\infty}y^{2}\varphi_{i}(y)dy=k^{2}_{i} \label{eq_3} \end{equation} and assume that, for any \textit{i}, k^{2}_{i} is negligible with respect to \sum^{n}_{1}k^{2}_{i}. Under these assumptions, Laplace's theorem holds which says that: The probability that inequalities \begin{equation} x\leq\sum^{n}_{1}y_{i}\leq x+dx \label{eq_4} \end{equation} hold at the same time is given by \begin{equation} F(x)dx =\frac{1}{\sqrt{2\pi \sum^{n}_{1}k^{2}_{i}}}e^{-\frac{x^{2}}{2\sum^{n}_{1}k^{2}_{i}}}dx\ . \label{eq_5} \end{equation} To demonstrate this, we indicate by r a number \leq n and let F(r,x)dx be the probability that the inequalities \begin{equation} x\leq\sum^{r}_{1}y_{i}\leq x+dx \label{eq_6} \end{equation} hold. Now if p is any value, let us look for the probability that inequalities \begin{equation} \sum^{r-1}_{1}y_{i}p>\sum^{r}_{1}y_{i} \label{eq_8} \end{equation} hold simultaneously is \[\int^{\infty}_{0}d\xi \,\, F(r-1,p+\xi)\int^{\infty}_{\xi}\varphi_{r}(y)dy\ .$ The difference of these two probabilities is obviously given by the difference between the probability that $\sum^{r}_{1}y_{i}>p$ and the probability that $\sum^{r-1}_{1}y_{i}>p$, that is by $\int^{\infty}_{p}F(r,x)dx-\int^{\infty}_{p}F(r-1,x)dx\ \ .$ Then we have $\int^{\infty}_{p}F(r,x)dx-\int^{\infty}_{p}F(r-1,x)dx =\int^{\infty}_{0}d\xi \,\, F(r-1,p-\xi)\int^{\infty}_{\xi}\varphi_{r}(y)dy$ $-\int^{\infty}_{0}d\xi F(r-1,p+\xi)\int^{\infty}_{\xi}\varphi_{r}(y)dy\ .$ On the right hand side we can reverse the integrations using the formulas $\int^{\infty}_{0}d\xi \int^{\infty}_{\xi}dy=\int^{\infty}_{0}dy\int^{y}_{0}d\xi\ \ \ ;\ \ \ \int^{\infty}_{0}d\xi \int^{-\xi}_{-\infty}dy=\int^{0}_{-\infty}dy\int^{-y}_{0}d\xi$ and it becomes, also changing $\xi$ into $-\xi$ in the second term $\int^{\infty}_{-\infty}\varphi_{r}(y)dy\int^{y}_{0}F(r-1,p-\xi)d\xi\ .$ We put, as an approximation $F(r-1,p-\xi)=F(r-1,p)-\xi\frac{\partial F(r-1,p)}{\partial p}\ .$ Thus the above expression becomes $F(r-1,p)\int^{\infty}_{-\infty}\varphi_{r}(y)dy\int^{y}_{0}d\xi-\frac{\partial F(r-1,p)}{\partial p}\int^{\infty}_{-\infty}\varphi_{r}(y)dy\int^{y}_{0}\xi d\xi$ $=F(r-1,p)\int^{\infty}_{-\infty}y\varphi_{r}(y)dy-\frac{1}{2}\frac{\partial F(r-1,p)}{\partial p}\int^{\infty}_{-\infty}y^{2}\varphi_{r}(y)dy$ namely, remembering (\ref{eq_2}) and (\ref{eq_3}): $-\frac{k^{2}_{r}}{2}\frac{\partial F(r-1,p)}{\partial p}.$ In this way we obtain the equality \begin{equation} \int^{\infty}_{p}F(r,x)dx-\int^{\infty}_{p}F(r-1,x)dx=-\frac{k^{2}_{r}}{2}\frac{\partial F(r-1,p)}{\partial p}. \label{eq_9} \end{equation} Differentiating it with respect to $p$ we obtain \begin{equation} -F(r,p)+F(r-1,p)=-\frac{k^{2}_{r}}{2}\frac{\partial^{2} F(r-1,p)}{\partial p^{2}}\ . \label{eq_10} \end{equation} Let us change in it $r-1$ into \textit{r}, \textit{p} into \textit{x}, and in our approximation, set $F(r+1,x)-F(r,x)=\frac{\partial}{\partial r}F(r,x)\ .$ Then (\ref{eq_10}) gives for $F(r,x)$ the differential equation \begin{equation} \frac{\partial}{\partial r}F(r,x) =-\frac{k^{2}_{r+1}}{2}\frac{\partial^{2}}{\partial x}F(r,x)\ . \label{eq_11} \end{equation} Changing \textit{r} into another variable \begin{equation} t=\int^{r+1}_{0}k^{2}_{i}di \label{eq_12} \end{equation} (\ref{eq_11}) becomes \begin{equation} \frac{\partial F}{\partial t}=\frac{1}{2}\frac{\partial^{2}F}{\partial x^{2}}\ . \label{eq_13} \end{equation} Then one has obviously the condition that for any \textit{t} \begin{equation} \int^{\infty}_{-\infty}Fdx=1 \label{eq_14} \end{equation} and that for $t=0$, \textit{F} has a nonvanishing value only when $\left|x\right|$ is infinitesimal. It is known that these conditions are more than sufficient to determine \textit{F}. They are satisfied by setting $F=\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^{2}}{2t}}\ .$ By giving to \textit{t} its value, which to our degree of approximation is $\sum^{r}_{1}k^{2}_{i}$, we find \begin{equation} F(r,x) =\frac{1}{\sqrt{2\pi \sum^{r}_{1}k^{2}_{i}}}e^{-\frac{x^{2}}{2\sum^{r}_{1}k^{2}_{i}}}\ . \label{eq_15} \end{equation} Then one obviously has $F(x)=F(n,x)$, and therefore \begin{center} $F(x)=\frac{1}{\sqrt{2\pi \sum^{n}_{1}k^{2}_{i}}}e^{-\frac{x^{2}}{2\sum^{n}_{1}k^{2}_{i}}}$\ \ \ \ \ \ \ \ \ \ \ \ Q.E.D. \end{center} \fermisection{2} Let us mantain the notations and the assumptions made at the beginning of the previous section and in addition assume that all $\varphi_{i}(y)$ are equal (as a consequence we will supress their index). Then let us indicate by $a$ any positive value. Thus we can state the following \begin{theorem}(REMOVE THEOREM NUMBER!!) The probability that at least one of the quantities $y_{1},y_{1}+y_{2}, y_{1}+y_{2}+y_{3},\ldots,\sum^{n}_{1}y_{n}$ exceeds \textit{a} is given by $\frac{2}{\sqrt{\pi}}\int^{\infty}_{\frac{a}{\sqrt{2nk^{2}}}}e^{-x^{2}}dx$ provided that \textit{a} is sufficiently large compared to \textit{k}. \end{theorem} In particular, if \textit{n} tends to infinity, such a probability tends to 1, i.e., to certainty. To demonstrate this, let us indicate by $F(r,x)dx(x\frac{\textnormal{M}}{\textnormal{R}}\] but we remark that $u=\sqrt{\frac{\textnormal{M}}{\textnormal{R}}}$ therefore the above inequality can be written: $\left(\textnormal{V}+\sqrt{\frac{\textnormal{M}}{\textnormal{R}}}\right)^{2}>\frac{2\textnormal{M}}{\textnormal{R}}$ \setcounter{footnote}{0} \renewcommand\thefootnote{\fnsymbol{footnote}} from which \footnote{Editor's Note: At this point, in the Fermi's manuscript there is a blank line which, obviously, would have contained the expansion of the square of the last formula (N.d.R.).} and reduces at the end to \begin{equation} \textnormal{V}>\left(\sqrt{2}-1\right)\sqrt{\frac{\textnormal{M}}{\textnormal{R}}}=\left(\sqrt{2}-1\right)u\ . \label{eq_40} \end{equation} Therefore we will assume that this inequality is certainly satisfied. Moreover, for some values of$\theta$, W must certainly be negative, otherwise the cometary orbit could not be elliptic; so it must be that: $\textnormal{V}^{2}+2u\textnormal{V}<\frac{\textnormal{M}}{\textnormal{R}}$ from which as above \begin{equation} \textnormal{V}>\left(\sqrt{2}+1\right)\sqrt{\frac{\textnormal{M}}{\textnormal{R}}}=\left(\sqrt{2}+1\right)u\ . \label{eq_41} \end{equation} Therefore let us assume that V satisfies (\ref{eq_40}) and (\ref{eq_41}) and indicate by$\theta_{0}$that particular value of$\theta$for which the comet's orbit is hyperbolic, i.e., one has W = 0, that is $\textnormal{V}^{2}+2u\textnormal{V}\cos\theta_{0}=\frac{\textnormal{M}}{\textnormal{R}}$ and then \begin{equation} \cos\theta_{0}=\frac{\frac{\textnormal{M}}{\textnormal{R}}-\textnormal{V}^{2}}{2u\textnormal{V}}=\frac{u^{2}-\textnormal{V}^{2}}{2u\textnormal{V}}\ . \label{eq_42} \end{equation} When$\theta$is greater than$\theta_{0}$, one has W$< 0$and then the comet describes an elliptic orbit; on the contrary, when$\theta$is less then$\theta_{0}$the orbit is hyperbolic. Now we will suppose that initially the orbit is elliptic and very stretched, so that$\theta$is very close to$\theta_{0}$, and precisely slightly greater. We call$\theta^{*}$this initial value. Whenever the comet goes beyond Jupiter's orbit$\theta$is changed of an amount$\psi$; the average of the squares of$\psi$depends indeed on$\theta$, as (\ref{eq_39}) shows, but since we have supposed that$\theta$remains always very close to$\theta_{0}$we can set \begin{equation} \bar{\psi^{2}}=\frac{4mh}{\pi \textnormal{R V}^{2}\sin\theta_{0}} \label{eq_43} \end{equation} if after a certain time$\theta$became <$\theta_{0}$the comet would become hyperbolic and would go away forever. Therefore we are in condition of being able to apply the theorem of \S 2. Then we must set$a=\theta^{*}-\theta_{0}$;$k^{2}=\frac{4mh}{\pi \textnormal{R V}^{2}\sin\theta_{0}}$. And the theorem we proved tells us that: The probability that the comet will be changed into a hyperbolic one after having crossed Jupiter's orbit$n$times is: \begin{equation} \frac{2}{\sqrt{\pi}}\int^{\infty}_{\frac{\theta^{*}-\theta_{0}}{\sqrt{\frac{8mhn}{\pi \textnormal{R V}^{2}\sin\theta_{0}}}}}e^{-x^{2}}dx \label{eq_44} \end{equation} and therefore tends to 1 when$n$tends to infinity. In the strict sense one could object that the above calculations would fail if the value of V were such that, when the orbit is parabolic, the comet took the same time as Jupiter to go from A to B, where A is the point where the comet enters Jupiter's orbit, and B the point where it goes out. In Figure~\ref{fig_1}, S is the Sun, AJB Jupiter's orbit, AKB the orbit of the comet. But it is easy to realize that this case certainly cannot happen if the comet describes its trajectory with direct motion. In fact, if \textit{v} is the absolute velocity in A of the comet in its parabolic orbit, one has $v^{2}=u^{2}+\textnormal{V}^{2}+2u\textnormal{V}\cos\theta_{0}$ and then from (\ref{eq_42}) $v^{2}=2u^{2}$ that is: \begin{equation} v>u\ . \label{eq_45} \end{equation} Now the velocity of the comet is not constant, but in whole tract AKB it is always greater than in the extremes A and B, thus inequality (\ref{eq_45}) holds true with all the more reason in the whole tract AKB. On the other hand, if the motion is direct one has that arc AKB is shorter than arc AJB, and since it is covered with even higher velocity it is certain that the comet will arrive at B before Jupiter. If on the contrary the motion of the comet were retrograde, and it described for instance the orbit AK'B' in the sense indicated by the arrow one would have $\textnormal{arc AK'B'}>\textnormal{arc AJB'}$ and then, though (\ref{eq_45}) still holds, it is evident that for a particular value of the parameter of the cometary orbit it can happen that the two heavenly bodies take the same time to go from A to B'; of course this can only happen for a particular value of V. \begin{figure}[ht] %\epsfxsize=10cm %width of figure - will enlarge/reduce the figures %\epsfbox{fig3.eps} %\figurebox{2cm}{3cm}{} %to have a box alone %\centerline{\epsfxsize=4.1in\epsfbox{Fig_1.eps}} \centerline{\epsfxsize=2.5in\epsfbox{fermi38bfig_1.eps}} \caption{\label{fig_1}} \end{figure} Now if this happened it could occur that the comet, elliptic at first, crossed Jupiter when passing through A and got changed into a parabolic one; but in this case it would meet Jupiter again when passing through B and could possibly have a new perturbation which would change it into an elliptic comet again. For this reason we consider this particular value of V ruled out from our calculations. \fermisection{7} Finally we want to consider the possibility that before being changed into a hyperbolic one the comet can crash into Jupiter and then be destroyed. What is the probability of this event? For this let us look first for the probability that the comet, crossing once Jupiter's orbit, collides with the planet. If we indicate by$\rho$the sum of the radii of Jupiter and the comet, to have the collision it is necessary that the periheliac distance of Jupiter from the comet, as calculated though the formulas of the Keplerian motion is smaller than$\rho$. Let$\delta$be this periheliac distance; from the formulas of \S 5 one finds that $\frac{1}{\delta}=\textnormal{A + B}$ and then from (\ref{eq_35}) and (\ref{eq_36}) $\frac{1}{\delta}=\frac{m}{\textnormal{V}^{2}b^{2}}+\frac{1}{b}\sqrt{1+\frac{m}{\textnormal{V}^{4}b^{2}}}$ If we want the collision to occur it must be that$\delta<\rho$and therefore $\frac{m}{\textnormal{V}^{2}b^{2}}+\frac{1}{b}\sqrt{1+\frac{m}{\textnormal{V}^{4}b^{2}}}>\frac{1}{\rho}$ by multiplying this inequality by the quantity, certainly positive $\rho\left(\frac{1}{b}\sqrt{1+\frac{m}{\textnormal{V}^{4}b^{2}}}>\frac{1}{\rho}-\frac{m}{\textnormal{V}^{2}b^{2}}\right)$ we find $\frac{\rho}{b^{2}}\frac{1}{b}\sqrt{1+\frac{m}{\textnormal{V}^{4}b^{2}}}>\frac{1}{\rho}-\frac{m}{\textnormal{V}^{2}b^{2}}$ and summing the last two inequalities $\left(\frac{2m}{\textnormal{V}^{2}}+\rho\right)\frac{1}{b^{2}}>\frac{1}{\rho}$ from which finally \begin{equation} \left|b\right|<\sqrt{\rho^{2}+\frac{2m\rho}{\textnormal{V}^{2}}}\ . \label{eq_46} \end{equation} We recall now that the probability that the value of \textit{b} lies between \textit{b} and \textit{b + db} is$\frac{db}{2\pi \textnormal{R}\sin\theta_{0}}$and that the probability \textit{p} that the collision occurs in only one crossing of Jupiter's orbit is given by \begin{equation} p=\frac{1}{\pi \textnormal{R}\sin\theta_{0}}\sqrt{\rho^{2}+\frac{2m\rho}{\textnormal{V}^{2}}}\ . \label{eq_47} \end{equation} We will assume \textit{p} to be very small, and this obviously is equivalent to considering Jupiter's radius negligible if compared with the radius of its orbit. Let us now look for the probability that the collision occurs at the \textit{n}-th time the comet crosses Jupiter's orbit. Therefore it is evidently necessary that the collision has not occurred before and the probability of this is obviously$(1-p)^{n-1}$, that is in our approximation $e^{-pn}\ .$ That the comet has not yet been changed into a hyperbolic one; and having supposed \textit{p}to be extremely small, remembering (\ref{eq_44}) and setting for the sake of brevity: $\frac{\theta^{*}-\theta_{0}}{\sqrt{\frac{8mh}{\pi \textnormal{RV}^{2}\sin\theta_{0}}}}=\textnormal{H}$ we can claim that the probability of this event is given by $1-\frac{2}{\sqrt{\pi}}\int^{\infty}_{\frac{\textnormal{H}}{\sqrt{n}}}e^{-x^{2}}dx=\frac{2}{\sqrt{\pi}}\int^{\frac{\textnormal{H}}{\sqrt{n}}}_{0}e^{-x^{2}}dx\ .$ And finally that the collision really occurs, for which we have the probability \textit{p}. After all the probability that the collision occurs the \textit{n}-th time is $\frac{2e^{-pn}p}{\sqrt{\pi}}\int^{\frac{\textnormal{H}}{\sqrt{n}}}_{0}e^{-x^{2}}dx$ and therefore the probability that the collision occurs at any time whatsoever will be the sum of the above expression from$n=1$to$n=\infty\$, or replacing the sum by an integral $\frac{2p}{\sqrt{\pi}}\int^{\infty}_{0}e^{-pn}dn\int^{\frac{\textnormal{H}}{\sqrt{n}}}_{0}e^{-x^{2}}dx\ .$ In this expression it is convenient to reverse the integration by the formula $\int^{\infty}_{0}dn\int^{\frac{\textnormal{H}}{\sqrt{n}}}_{0}dx =\int^{\infty}_{0}dx\int^{\frac{\textnormal{H}}{x^{2}}}_{0}dn$ and in this way one finds for the desired probability the expression: $\frac{2p}{\sqrt{\pi}}\int^{\infty}_{0}e^{-x^{2}}dx\int^{\frac{\textnormal{H}}{x^{2}}}_{0}e^{-pn}dn=\frac{2}{\sqrt{\pi}}\int^{\infty}_{0}e^{-x^{2}}\left(1-e^{-\frac{p\textnormal{H}}{x^{2}}}\right)dx$ $=1-\frac{2}{\sqrt{\pi}}\int^{\infty}_{0}e^{-x^{2}-\frac{p\textnormal{H}}{x^{2}}}dx=1-e^{-2\sqrt{p\textnormal{H}}}\ .$ The probability that the collision never occurs is therefore: $e^{-2\sqrt{p\textnormal{H}}}\ .$ \end{document}