From
The solution is found when the difference is 0 (10 iterations approx.)
From Scipy: Non-linear solvers
fsolve
is used to estimate
The dataset Cumulative Distribution is:
Note: Analytical tests should be performed to validate
And
-
$X\sim\beta(\alpha, \beta)$ with$f(x_{i}, f_{x_{i}}):$
x | f(x) | F(x) |
---|---|---|
40.00 | inf | 0.000000 |
40.01 | 1.040696 | 0.019154 |
40.02 | 0.758528 | 0.027915 |
40.03 | 0.630507 | 0.034797 |
40.04 | 0.553063 | 0.040688 |
... | ... | ... |
48.95 | 0.247013 | 0.981838 |
48.96 | 0.265219 | 0.984395 |
48.97 | 0.290729 | 0.987167 |
48.98 | 0.330971 | 0.990257 |
48.99 | 0.413234 | 0.993916 |
-
$X\sim\text{T}(a,b,c)$ :
x | f(x) | F(x) |
---|---|---|
0 | 40.000000 | 0.000000 |
1 | 40.090909 | 0.005102 |
2 | 40.181818 | 0.010203 |
3 | 40.272727 | 0.015305 |
4 | 40.363636 | 0.020406 |
... | ... | ... |
95 | 48.636364 | 0.997085 |
96 | 48.727273 | 0.998360 |
97 | 48.818182 | 0.999271 |
98 | 48.909091 | 0.999818 |
99 | 49.000000 | 1.000000 |
- Montecarlo Estimation:
if
and it follows a uniform prob. density function np.random.uniform
.
Nevertheless if the distribution isn't known it must be obtained from data or modelled with another method.
From where we've got by the definition the expectancy:
Then, the mean of
if
But Montecarlo estimations for
Newton Raphson
Taylor-Series
Jacobian
scipy.stats.beta
scipy.stats.gamma
scipy.stats.triang