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

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