By Gerhard Bohm, Günter Zech

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**Sample text**

3. Probability density and distribution function of a continuous distribution. 2) where the parameter2 λ > 0, the decay constant, is the inverse of the mean lifetime τ = 1/λ. The probability density and the distribution function t F (t) = −∞ f (t′ )dt′ = 1 − e−λt are shown in Fig. 4. The probability of observing a lifetime longer than τ is P {t > τ } = F (∞) − F (τ ) = e−1 . Example 8. Probability density of the normal distribution An oxygen atom is drifting in argon gas, driven by thermal scattering.

We deﬁne the distribution function F (x1 , . . , xN ) as the probability to ﬁnd values of the variates smaller than x1 , . . f. F (x1 , . . , xN ) = P {(x′1 < x) ∩ · · · ∩ (x′N < xN )} , f (x1 , . . , xN ) = ∂N F . ∂x1 , . . , ∂xN Often it is convenient to use the vector notation, F (x), f (x) with x = {x1 , x2 , . . , xN } . These variate vectors can be represented as points in an N -dimensional space. f. f (x) can also be deﬁned directly, without reference to the distribution function F (x), as the density of points at the location x, by setting dx1 dx1 ≤ x′1 ≤ x1 + )∩ ··· 2 2 dxN dxN ≤ x′N ≤ xN + )} · · · ∩(xN − 2 2 f (x1 , .

The relation P {x1 < x < x2 } = P {u1 < u < u2 } with u1 = u(x1 ), u2 = u(x2 ) has to hold, and therefore x2 P {x1 < x < x2 } = x1 u2 = f (x′ ) dx′ g(u′ ) du′ . 29) |g(u)du| = |f (x)dx| , dx . g(u) = f (x) du Taking the absolute value guarantees the positivity of the probability density. 29), we ﬁnd numerical equality of the cumulative distribution functions, F (x) = G(u(x)). If u(x) is not a monotone function, then, contrary to the above assumption, x(u) is not a unique function (Fig. 9) and we have to sum over the contributions of the various branches of the inverse function: g(u) = f (x) dx du + f (x) branch1 dx du branch2 + ··· .