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A rock is dropped off a cliff of height h. As it falls a million pics are taken (at random intervals). On each picture it is measured the distance the rock has fallen. What is the average of these distances?

In order to solve this problem the probability density function $\rho(x)dx$ must be known.

The probability of taking a picture in the interval dt is $\frac{dt}{T}$, where $T = \sqrt{\frac{2h}{g}}$. The rock is initially at rest thus: $v = gt$

Then, theoretically the probability density function can be deduced from the probability of taking a picture in the interval dt:

$$\frac{dt}{T} = \frac{dx}{gt}\sqrt{\frac{g}{2h}}$$

However, the probability density function I am supposed to get is:

$$\rho(x) = \frac{1}{2\sqrt{hx}}$$

While what I got:

$$\rho(x) = \frac{1}{gt}\sqrt{\frac{g}{2h}}$$

I think my mistake is algebraic.

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  • $\begingroup$ What is the relationship between x and t? Also total probability has to be 1. $\endgroup$ – herb steinberg Aug 7 '18 at 23:47
  • $\begingroup$ @herbsteinberg v=x/t. Absolutely, that is why I know I am wrong. If you compute the integral from h to 0 of my result you do not obtain 1. If you do so on the right answer you get 1 $\endgroup$ – JD_PM Aug 7 '18 at 23:54
  • $\begingroup$ You asserted the probability is $\frac{dt}{T}$, but you need a normalization factor. $\endgroup$ – herb steinberg Aug 8 '18 at 0:00
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You have $f_t(\tau)=\dfrac{\mathbf 1_{\tau\in(0;T)}}{T}$ and $x=\tfrac 12 g~\tau^2$ so also $T=\sqrt{2h/g}$

So indeed you should get, via the fundamental principle of calculus and the chain rule:

$$\begin{align}f_X(x) &= \left\lvert\dfrac{\mathsf d F_t(\surd(2x/g))}{\mathsf d x\qquad\qquad~~} \right\rvert \\&=\left\lvert\dfrac{\mathsf d ~~}{\mathsf d x} \int_{-\infty}^{\surd(2x/g)} f_t(\tau)~\mathsf d \tau \right\rvert\\ &= \left\lvert\dfrac{\mathsf d \sqrt{2x/g~}}{\mathsf d x}\right\rvert f_t(\surd(2x/g))\\ &= \left\lvert\dfrac{\mathsf d \sqrt{2x/g~}}{\mathsf d x}\right\rvert \dfrac{\mathbf 1_{x\in(0;h)}}{\surd(2h/g)}\\& ~~\vdots\\&=\dfrac{\mathbf 1_{x~\in~(0;h)}}{2\sqrt {hx}}\end{align}$$

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  • $\begingroup$ Sorry Graham but I do not understand your definition of $fx(x)$. What I did was just isolate $dt$ and $T$. Could you shed some light on it? $\endgroup$ – JD_PM Aug 8 '18 at 0:39
  • $\begingroup$ You cannot just 'isolate' $\mathsf d t$ and $T$. The probability density function is the unsigned derivative of the cummulative distribution function. $\endgroup$ – Graham Kemp Aug 8 '18 at 0:47
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$\frac{dt}{T} = \frac{dx}{gt}\sqrt{\frac{g}{2h}} = dx\sqrt{\frac{g\times2}{g^2t^22h\times 2}} = \frac{dx}{2\sqrt{h}}\sqrt{\frac{2\times\frac{1}{2}}{gt^2\times\frac{1}{2}}} = \frac{1}{2\sqrt{hx}}$

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