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|newpage| Cumulative distribution function (cdf) ======================================================== Calculating the cdf from the pdf ------------------------------------------------------------------------------- .. method:: ctx.cdf_from_pdf(x, cf) where ``ctx`` is ``dec``, ``mpm``, or ``gmp``. Returns `\text{cdf}_X(x)`, the cumulative distribution function (cdf) of a random variable `X`: .. math:: \text{cdf}_X(x) = \int_{0}^{x} \text{pdf}_X(x) \mathrm{d}t . Using this method can be a viable option when the pdf, but not the cdf, is available in closed form. Calculating the cdf from the pmf ------------------------------------------------------------------------------- .. method:: ctx.cdf_from_pmf(x, cf) where ``ctx`` is ``dec``, ``mpm``, or ``gmp``. If `X` is a purely discrete random variable, then it attains values `x_{1},x_{2},\ldots` with probability `p_{i}=p(x_{i})`, and the CDF of `X` will be discontinuous at the points `x_{i}`: .. math:: F_{X}(x)=\operatorname {P} (X\leq x)=\sum _{x_{i}\leq x}\operatorname {P} (X=x_{i})=\sum _{x_{i}\leq x}p(x_{i}). .. _rst_gil_pelaez_cdf: Calculating the cdf from the characteristic function ------------------------------------------------------------------------------- .. method:: ctx.cdf_from_cf_continuous(x, cf) where ``ctx`` is ``dec``, ``mpm``, or ``gmp``. Calculates the cdf from the characteristic function using the procedure of Gil-Pelaez. Assuming that the characteristic function is absolutely integrable over `(-\infty, \infty)`, Gil-Pelaez derived the following inversion formula which requires integration of a real-valued function only. In particular, .. math:: \text{cdf}_X(x) = \frac{1}{2} - \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{e^{-itx} C_X(t) - e^{itx} C_X(t)}{it} \mathrm{d} t = \frac{1}{2} - \frac{1}{\pi} \int_{0}^{\infty} \Im \left ( \frac{ e^{-itx} C_X(t)}{t} \right ) \mathrm{d}t. where `\Im (z)` denotes the imaginary part of `z`. We also have .. math:: \int_{0}^{\infty} \Im \left ( \frac{ e^{-itx} C_X(t)}{t} \right ) \mathrm{d}t = \int_{0}^{\infty} \Im \left ( \frac{C_X(t)}{t} \right ) \cos(t x) \mathrm{d}t - \int_{0}^{\infty} \Re \left ( \frac{C_X(t)}{t} \right ) \sin(t x) \mathrm{d}t. Using the right-hand side of this equation allows for efficient use of the quadrature formula of Fillon, with the half-period `\omega = \pi/x`. The python code is currently in Charfun.py The following code provides a test-suite for the numerical inversion of the characteristic function: .. code-block:: python class tests_charfunc(rv_cont): def __init__(self, rv2, x = 5, a = 0, b = 2): cdf_value = rv2.cdf(x) print ("rv2.cdf(x): ", cdf_value) rv2.set_x(x) plot(rv2.gil_pelaez_imag, [a, b], points=200) print rv2.set_x(x) plot(rv2.gil_pelaez_cos, [a, b], points=200) print rv2.set_x(x) plot(rv2.gil_pelaez_sin, [a, b], points=200) print rv2.set_x(x) I0 = quad(rv2.gil_pelaez_imag, [0, +inf]) print("Integral: ", I0) result0 = 0.5 - I0/pi print("result0:", result0 ) print("diff0:", result0 - cdf_value) rv2.set_x(x) I1 =quadosc(rv2.gil_pelaez_cos, [0, inf], period=1*pi/x) # half period print("I1:", I1 ) rv2.set_x(x) I2 =quadosc(rv2.gil_pelaez_sin, [0, inf], period=1*pi/x) # half period print("I2:", I2 ) I3 = I1 + I2 print("I3:", I3 ) print("Int diff:", I3 - I0) result3 = 0.5 - I3/pi print("result3:", result3 ) print("diff3:", result3 - cdf_value) **Example: non-central chi-squared distribution** The following code shows the difference between using the generic integration (error: diff0: -6.1218939724378892896e-6) and quadrature for oscillatory functions (error: diff3: 8.4703294725430033907e-22). .. code-block:: python mp.dps = 20 print() print ("Hello mpDistributions local ! ") print() a = 0.0 b = 2 n = mpf("5") x = mpf("10") rv2 = mpr().chisquare(n) tests_charfunc(rv2, x, a, b) This produces the following output (plots are to be added): .. parsed-literal:: Hello mpDistributions local ! rv2.cdf(x): 0.92476475385348782128 Integral: -1.334418597712864291 result0: 0.92475863195951538339 diff0: -6.1218939724378892896e-6 I1: 0.11817924829245123086 I2: -1.4526170785024453884 I3: -1.3344378302099941575 Int diff: -0.000019232497129866511475 result3: 0.92476475385348782128 diff3: 8.4703294725430033907e-22 .. _rst_cdf_from_cf_lattice: Calculating the cdf and sf from the characteristic function (lattice distribution) -------------------------------------------------------------------------------------------- .. method:: ctx.cdf_from_cf_lattice(x, cf) where ``ctx`` is ``dec``, ``mpm``, or ``gmp``. The corresponding inversion formula for discrete distributions on the nonnegative integers is .. math:: \text{cdf}(x) = \frac{1}{\pi} \int_{0}^{\pi} \Re \left( C_X(t) \sum_{z=0}^x e^{-itz} \right) \mathrm{d}t. .. math:: \text{sf}(x) = 1-\text{cdf}(x) = \frac{1}{\pi} \int_{0}^{\pi} \Re \left( C_X(t) \sum_{z=x+1}^{N(N+1)/2} e^{-itz} \right) \mathrm{d}t. Calculating the cdf from the factorial moments (lattice distributions) --------------------------------------------------------------------------------------------- The following relationships hold between the probabilities and the factorial moments in the case of a discrete distribution: .. math:: \text{Pr}[X=x] = \sum_{j \ge x} (-1)^{x+j} \binom{j}{x} \frac{\mu'_{[j]}}{j!} = \sum_{r \ge 0} (-1)^{r} \frac{\mu'_{[x+r]}}{x!r!} and .. math:: \sum_{i \ge x} \text{Pr}[X=i] = \sum_{j \ge x} (-1)^{x+j} \binom{j-1}{x-1} \frac{\mu'_{[j]}}{j!} See also Johnson(2005), page 59.