Probability density function for a given problem

[Maxdenis92]

https://www.quora.com/How-can-I-find-the-probability-density-function-of-a-continuous-random-variable-in-a-given-problem/answer/Maxime-Denis-2 How can I find the probability density function of a continuous random variable in a given problem?

A mass m swings at the end of a rope (of length L) such as a pendulum above a pond (of half width 2 m) according to a planar oscillation (of amplitude 5m) as depicted in FIG. The branch can break at any time during the cycle. Calculate the probability that the mass will fall to the water as a result of this random break. (Neglect horizontal speed during fall).

I know how to calculate the probability (which is approximately 0.26), but I would like to find a probability density function and the corresponding cumulative distribution function. Excuse-me for my English mistakes. English is my second language.

 

[Charles Link]

I think this one really belongs in the homework section. Please post homeworks in the homework section in the future and fill out the homework template. $\\$ In any case, this one can be written as $x(t)=A \sin(\omega t)$ where $\omega=\frac {2 \pi}{T}$. Basically the density function of the position vs. time can be found by solving for $t$ : $\, t=\frac{1}{\omega} \sin^{-1}(x/A)$ and finding $f(x) \Delta x= \Delta t/(T/2)$. $f(x)$ is the probability density function for the probability that $x<X \leq x+\Delta x$. It is found by taking a derivative which will be found to diverge at $x=A$ and $x=-A$, but the integral from $x=-A$ to $x$ where $-A<x<A$ should be found to converge. For $x<-A$ and $x>A$, $f(x)=0$. To find the probability it lands in the pond, you need to integrate the density function over the limits of the pond. $\\$ The forum rules requite the homework helper doesn't give the solution to the problem. Because this one is semi-difficult, I have provided more info than I normally would. I believe what I have provided is correct, and hopefully this is helpful. $\\$ Editing: Note: $T/2$ is the proper normalization factor. The time the pendulum spends at the different locations is thereby found from this normalized density function. It is only necessary to consider a half cycle of the pendulum, from $x=-A$ to $x=A$.

 

[Maxdenis92]

Hello Charles Link. Thank you for your answer. This is not an homework. I just want to know what is the density function and the cumulative distribution function. I know that f(x) must be positive, but f(x) = t(x) = (1/w)arcin(x/A) is not always positive. I don't see what I have to integrate to get the cumulative distribution function. I did it in a probability course, but there was no context. Thank you.

 

[Charles Link]

For $0<x<A$ it works ok. The probability density function $f(x)$ is symmetric about $x=0$. For $x<0$, let $f(x)=f(-x)$. $\\$ Editing: $t=\frac{1}{\omega} \sin^{-1}(\frac{x}{A})$, so that $f(x) \Delta x=\Delta t/(T/2)=(\frac{1}{\pi A}) \frac{1}{\sqrt{1-(\frac{x}{A})^2 } } \Delta x$. The function $f(x)$ has the necessary properties. (Note: There's actually a +/- sign on the derivative, but it works to just keep the + sign. We are considering $-T/4<t<T/4$ with $-A<x<A$ which is a half cycle of the pendulum). $\\$ For the problem at hand $A=5$ and the half width of the pond is $2$ , so to get the probability $P=\int\limits_{-2}^{2} f(x) \, dx$ that it lands in the pond. Meanwhile $T=2 \pi \sqrt{L/g}$ where $g=9.8 \, m/sec^2$, and $\omega=2 \pi /T$. But this is extra information.

 

[Charles Link]

Note: I made a correction to the above: $\omega T=2 \pi$, thereby the $1/(\pi A)$ term in $f(x)$. $\\$ The integral $P=(\frac{1}{ 5 \pi }) \int\limits_{-2}^{2} \frac{1}{\sqrt{1-(x/5)^2}} \, dx$ can be readily performed, and the result is $P= (\frac{2}{\pi}) \sin^{-1}(2/5) \approx .26$. $\\$ Had the half width of the pond been equal to 5, then $P=(\frac{2}{\pi}) \sin^{-1}(5/5)=(\frac{2}{\pi})(\frac{\pi}{2})=1$.