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# statsmodels ols get_prediction Implementation. Unfortunately, our specification allows us to calculate the prediction of the log of $$Y$$, $$\widehat{\log(Y)}$$. I try to import matplotlib.pyplt in Pycharm console import matplotlib.pyplot as plt Then in return I get: Traceback (most recent call last): File "D:\Program Files\Anaconda2\lib\site-packages\IPython\core\interactiveshell.py", line 2881, in run_ Dataset Description 2. Unemployment RatePlease note that you will have to validate that several assumptions are met before you apply linear regression models. \end{aligned} which we can rewrite as a log-linear model: \], $$\widehat{\sigma}^2 = \dfrac{1}{N-2} \sum_{i = 1}^N \widehat{\epsilon}_i^2$$, $$\text{se}(\widetilde{e}_i) = \sqrt{\widehat{\mathbb{V}{\rm ar}} (\widetilde{e}_i)}$$, $$\widehat{\mathbb{V}{\rm ar}} (\widetilde{\boldsymbol{e}})$$, Y = \beta_0 + \beta_1 X + \epsilon \mathbb{V}{\rm ar}\left( \widetilde{\boldsymbol{e}} \right) &= We again highlight that $$\widetilde{\boldsymbol{\varepsilon}}$$ are shocks in $$\widetilde{\mathbf{Y}}$$, which is some other realization from the DGP that is different from $$\mathbf{Y}$$ (which has shocks $$\boldsymbol{\varepsilon}$$, and was used when estimating parameters via OLS). Fit an OLS regression model on the counts data set to find the value of α that is used in the variance function of the NB2 model (refer to equation of the variance function above). Then, the $$100 \cdot (1 - \alpha) \%$$ prediction interval can be calculated as: \[ \begin{aligned} We can estimate the systematic component using the OLS estimated parameters: \], $$\left[ \exp\left(\widehat{\log(Y)} \pm t_c \cdot \text{se}(\widetilde{e}_i) \right)\right]$$, On the other hand, in smaller samples $$\widehat{Y}$$ performs better than $$\widehat{Y}_{c}$$. This is also known as the standard error of the forecast. Prediction intervals must account for both: (i) the uncertainty of the population mean; (ii) the randomness (i.e.Â scatter) of the data. To be included after running your script: This should give the same results as SAS, http://jpktd.blogspot.ca/2012/01/nice-thing-about-seeing-zeros.html. \[ \end{aligned} I need the confidence and prediction intervals for all points, to do a plot. We can defined the forecast error as &= \sigma^2 \mathbf{I} + \widetilde{\mathbf{X}} \sigma^2 \left( \mathbf{X}^\top \mathbf{X}\right)^{-1} \widetilde{\mathbf{X}}^\top \\ Using the conditional moment properties, we can rewrite $$\mathbb{E} \left[ (Y - g(\mathbf{X}))^2 \right]$$ as: Another way to look at it is that a prediction interval is the confidence interval for an observation (as opposed to the mean) which includes and estimate of the error. &= 0 \begin{aligned} \mathbb{E} \left[ (Y - \mathbb{E} [Y|\mathbf{X}])^2 \right] = \mathbb{E}\left[ \mathbb{V}{\rm ar} (Y | X) \right]. \mathbf{Y} = \mathbb{E}\left(\mathbf{Y} | \mathbf{X} \right) \widetilde{\boldsymbol{e}} = \widetilde{\mathbf{Y}} - \widehat{\mathbf{Y}} = \widetilde{\mathbf{X}} \boldsymbol{\beta} + \widetilde{\boldsymbol{\varepsilon}} - \widetilde{\mathbf{X}} \widehat{\boldsymbol{\beta}} &= \sigma^2 \left( \mathbf{I} + \widetilde{\mathbf{X}} \left( \mathbf{X}^\top \mathbf{X}\right)^{-1} \widetilde{\mathbf{X}}^\top\right) $Since our best guess for predicting $$\boldsymbol{Y}$$ is $$\widehat{\mathbf{Y}} = \mathbb{E} (\boldsymbol{Y}|\boldsymbol{X})$$ - both the confidence interval and the prediction interval will be centered around $$\widetilde{\mathbf{X}} \widehat{\boldsymbol{\beta}}$$ but the prediction interval will be wider than the confidence interval. Some of the models and results classes have now a get_prediction method that provides additional information including prediction intervals and/or confidence intervals for the predicted mean. ... nb2_predictions = nb2_training_results. From the distribution of the dependent variable: &= \exp(\beta_0 + \beta_1 X) \cdot \exp(\epsilon)\\$, $Prediction interval is the confidence interval for an observation and includes the estimate of the error. Y = \exp(\beta_0 + \beta_1 X + \epsilon) Author: josef-pktd License: BSD """ import numpy as np from scipy import stats import scikits.statsmodels.api as sm from scikits.statsmodels.tsa.stattools import acf, adfuller from scikits.statsmodels.tsa.tsatools import lagmat #get the old signature back so the examples work def unitroot_adf(x, maxlag=None, trendorder=0, autolag='AIC', store=False): return adfuller(x, … 返回 下载statsmodels： 单独下载arima_model.py源代码 - 下载整个statsmodels源代码 - 类型：.py文件 # Note: The information criteria add 1 to the number of parameters # whenever the model has an AR or MA term since, in principle, For anyone with the same question: As far as I understand, obs_ci_lower and obs_ci_upper from results.get_prediction(new_x).summary_frame(alpha=alpha) is what you're looking for. \log(Y) = \beta_0 + \beta_1 X + \epsilon$ \], $$\mathbb{E}\left(\widetilde{Y} | \widetilde{X} \right) = \beta_0 + \beta_1 \widetilde{X}$$, , $$g(\mathbf{X}) = \mathbb{E} [Y|\mathbf{X}]$$, $\text{argmin}_{g(\mathbf{X})} \mathbb{E} \left[ (Y - g(\mathbf{X}))^2 \right].$ &= \mathbb{E} \left[ (Y - \mathbb{E} [Y|\mathbf{X}])^2 + 2(Y - \mathbb{E} [Y|\mathbf{X}])(\mathbb{E} [Y|\mathbf{X}] - g(\mathbf{X})) + (\mathbb{E} [Y|\mathbf{X}] - g(\mathbf{X}))^2 \right] \\ Therefore we can use the properties of the log-normal distribution to derive an alternative corrected prediction of the log-linear model: Confidence intervals are there for OLS but the access is a bit clumsy. &= \mathbb{V}{\rm ar}\left( \widetilde{\mathbf{Y}} \right) + \mathbb{V}{\rm ar}\left( \widehat{\mathbf{Y}} \right)\\ \], Assume that the data really are randomly sampled from a Gaussian distribution. \[ To generate prediction intervals in Scikit-Learn, we’ll use the Gradient Boosting Regressor, working from this example in the docs. We’ll see how to perform this regression using the Python statsmodels library. Develop Model 4. \widehat{Y}_i \pm t_{(1 - \alpha/2, N-2)} \cdot \text{se}(\widetilde{e}_i) We have examined model specification, parameter estimation and interpretation techniques. In the following example, we will use multiple linear regression to predict the stock index price (i.e., the dependent variable) of a fictitious economy by using 2 independent/input variables: 1. &= 0 Python statsmodels get_prediction function formula. \end{aligned} Thus, $$g(\mathbf{X}) = \mathbb{E} [Y|\mathbf{X}]$$ is the best predictor of $$Y$$. Assume that the best predictor of $$Y$$ (a single value), given $$\mathbf{X}$$ is some function $$g(\cdot)$$, which minimizes the expected squared error: If you sample the data many times, and calculate a confidence interval of the mean from each sample, youâd expect about $$95\%$$ of those intervals to include the true value of the population mean. (Actually, the confidence interval for the fitted values is hiding inside the summary_table of influence_outlier, but I need to verify this.). ... Confidence intervals are there for OLS … Taking $$g(\mathbf{X}) = \mathbb{E} [Y|\mathbf{X}]$$ minimizes the above equality to the expectation of the conditional variance of $$Y$$ given $$\mathbf{X}$$: Y &= \exp(\beta_0 + \beta_1 X + \epsilon) \\ For example, the code below will train an AR(6) model on the entire Female Births dataset and save it using the built-in save() function, which will essentially pickle the AutoRegResults object. Let $$\text{se}(\widetilde{e}_i) = \sqrt{\widehat{\mathbb{V}{\rm ar}} (\widetilde{e}_i)}$$ be the square root of the corresponding $$i$$-th diagonal element of $$\widehat{\mathbb{V}{\rm ar}} (\widetilde{\boldsymbol{e}})$$. \[ \[ Parameters: exog (array-like, optional) – The values for which you want to predict. \widehat{Y} = \exp \left(\widehat{\log(Y)} \right) = \exp \left(\widehat{\beta}_0 + \widehat{\beta}_1 X\right) ... #add a derived column called 'AUX_OLS_DEP' to the pandas Data Frame. In order to do that we assume that the true DGP process remains the same for $$\widetilde{Y}$$., $$\widetilde{\mathbf{X}} \boldsymbol{\beta}$$, $Variable: brozek: R-squared: 0.749: Model: OLS: Adj. where: The expected value of the random component is zero. Simple ANOVA Examples¶ Introduction¶. \[ \widehat{\mathbf{Y}} = \widehat{\mathbb{E}}\left(\widetilde{\mathbf{Y}} | \widetilde{\mathbf{X}} \right)= \widetilde{\mathbf{X}} \widehat{\boldsymbol{\beta}} However, usually we are not only interested in identifying and quantifying the independent variable effects on the dependent variable, but we also want to predict the (unknown) value of $$Y$$ for any value of $$X$$. A confidence interval gives a range for $$\mathbb{E} (\boldsymbol{Y}|\boldsymbol{X})$$, whereas a prediction interval gives a range for $$\boldsymbol{Y}$$ itself.$, , \mathbb{E} \left[ (Y - \mathbb{E} [Y|\mathbf{X}])^2 \right] = \mathbb{E}\left[ \mathbb{V}{\rm ar} (Y | X) \right]. the prediction is comprised of the systematic and the random components, but they are multiplicative, rather than additive. For larger samples sizes $$\widehat{Y}_{c}$$ is closer to the true mean than $$\widehat{Y}$$. The same ideas apply when we examine a log-log model. We begin by outlining the main properties of the conditional moments, which will be useful (assume that $$X$$ and $$Y$$ are random variables): For simplicity, assume that we are interested in the prediction of $$\mathbf{Y}$$ via the conditional expectation: \end{aligned} Furthermore, this correction assumes that the errors have a normal distribution (i.e.Â that (UR.4) holds). Each of the examples shown here is made available as an IPython Notebook and as a plain python script on the statsmodels github repository. \[ \widehat{Y}_{c} = \widehat{\mathbb{E}}(Y|X) \cdot \exp(\widehat{\sigma}^2/2) = \widehat{Y}\cdot \exp(\widehat{\sigma}^2/2) The Python statsmodels library also supports the NB2 model as part of the Generalized Linear Model class that it offers. We will show that, in general, the conditional expectation is the best predictor of $$\mathbf{Y}$$. statsmodels v0.13.0.dev0 (+127) Prediction (out of sample) Type to start searching statsmodels Examples; statsmodels v0.13.0.dev0 (+127) ... OLS Adj. \end{aligned} The key point is that the confidence interval tells you about the likely location of the true population parameter. &=\mathbb{E} \left[ \mathbb{E}\left((Y - \mathbb{E} [Y|\mathbf{X}])^2 | \mathbf{X}\right)\right] + \mathbb{E} \left[ 2(\mathbb{E} [Y|\mathbf{X}] - g(\mathbf{X}))\mathbb{E}\left[Y - \mathbb{E} [Y|\mathbf{X}] |\mathbf{X}\right] + \mathbb{E} \left[ (\mathbb{E} [Y|\mathbf{X}] - g(\mathbf{X}))^2 | \mathbf{X}\right] \right] \\ We estimate the model via OLS and calculate the predicted values $$\widehat{\log(Y)}$$: We can plot $$\widehat{\log(Y)}$$ along with their prediction intervals: Finally, we take the exponent of $$\widehat{\log(Y)}$$ and the prediction interval to get the predicted value and $$95\%$$ prediction interval for $$\widehat{Y}$$: Alternatively, notice that for the log-linear (and similarly for the log-log) model: Next, we will estimate the coefficients and their standard errors: For simplicity, assume that we will predict $$Y$$ for the existing values of $$X$$: Just like for the confidence intervals, we can get the prediction intervals from the built-in functions: Confidence intervals tell you about how well you have determined the mean. $&= \mathbb{E}(Y|X)\cdot \exp(\epsilon) \mathbf{Y} | \mathbf{X} \sim \mathcal{N} \left(\mathbf{X} \boldsymbol{\beta},\ \sigma^2 \mathbf{I} \right) We know that the true observation $$\widetilde{\mathbf{Y}}$$ will vary with mean $$\widetilde{\mathbf{X}} \boldsymbol{\beta}$$ and variance $$\sigma^2 \mathbf{I}$$. For the time series data set, we’ll use weather data downloaded from NOAA‘s website. I do this linear regression with StatsModels: My questions are, iv_l and iv_u are the upper and lower confidence intervals or prediction intervals? Because $$\exp(0) = 1 \leq \exp(\widehat{\sigma}^2/2)$$, the corrected predictor will always be larger than the natural predictor: $$\widehat{Y}_c \geq \widehat{Y}$$.$, $Thanks to Josef Perktold at StatsModels for assistance with the quantile regression code, ... OLS Regression Results ... (quantiles, res_all): # get prediction for the model and plot # here we use a dict which works the same way as the df in ols plt. Parámetros: params: array-like .$. \] statsmodels.regression.linear_model.OLSResults¶ class statsmodels.regression.linear_model.OLSResults (model, params, normalized_cov_params=None, scale=1.0, cov_type='nonrobust', cov_kwds=None, use_t=None, **kwargs) [source] ¶. \begin{aligned} \mathbf{Y} = \mathbb{E}\left(\mathbf{Y} | \mathbf{X} \right) \], get_prediction (X_test) #print out the predictions: What formula does this function use after computing a simple linear regression ... but I cannot find them in the index/module page. \widehat{\mathbf{Y}} = \widehat{\mathbb{E}}\left(\widetilde{\mathbf{Y}} | \widetilde{\mathbf{X}} \right)= \widetilde{\mathbf{X}} \widehat{\boldsymbol{\beta}} \left[ \exp\left(\widehat{\log(Y)} - t_c \cdot \text{se}(\widetilde{e}_i) \right);\quad \exp\left(\widehat{\log(Y)} + t_c \cdot \text{se}(\widetilde{e}_i) \right)\right] So, a prediction interval is always wider than a confidence interval. ; transform (bool, optional) – If the model was fit via a formula, do you want to pass exog through the formula.Default is True. ALlow Series to be used as exog in predict closes statsmodels#6509 bashtage mentioned this issue Jul 2, 2020 BUG: Allow Series as exog in predict #6847 class statsmodels.sandbox.regression.gmm.IVRegressionResults(model, params, normalized_cov_params=None, scale=1.0, cov_type='nonrobust', cov_kwds=None, use_t=None, **kwargs) [source] Results class for for an OLS model. I think, confidence interval for the mean prediction is not yet available in statsmodels. # Let's calculate the mean resposne (i.e. Because, if $$\epsilon \sim \mathcal{N}(\mu, \sigma^2)$$, then $$\mathbb{E}(\exp(\epsilon)) = \exp(\mu + \sigma^2/2)$$ and $$\mathbb{V}{\rm ar}(\epsilon) = \left[ \exp(\sigma^2) - 1 \right] \exp(2 \mu + \sigma^2)$$. Assume that the data really are randomly sampled from a Gaussian distribution. \begin{aligned} This algorithm’s calculation of the MLE (Maximum-Likelihood Estimate) means one value for each parameter estimated, i.e. 3 elementos iterables, con el número de parámetros AR, MA y exógenos, incluida la tendencia $\[ Proper prediction methods for statsmodels are on the TODO list. Having estimated the log-linear model we are interested in the predicted value $$\widehat{Y}$$.$. &= \mathbb{E}\left[ \mathbb{V}{\rm ar} (Y | X) \right] + \mathbb{E} \left[ (\mathbb{E} [Y|\mathbf{X}] - g(\mathbf{X}))^2\right]. \mathbb{E} \left[ (Y - g(\mathbf{X}))^2 \right] &= \mathbb{E} \left[ (Y + \mathbb{E} [Y|\mathbf{X}] - \mathbb{E} [Y|\mathbf{X}] - g(\mathbf{X}))^2 \right] \\ update see the second answer which is more recent. \mathbb{C}{\rm ov} (\widetilde{\mathbf{Y}}, \widehat{\mathbf{Y}}) &= \mathbb{C}{\rm ov} (\widetilde{\mathbf{X}} \boldsymbol{\beta} + \widetilde{\boldsymbol{\varepsilon}}, \widetilde{\mathbf{X}} \widehat{\boldsymbol{\beta}})\\