## statsmodels ols prediction interval

December 12th, 2020

\], $We have examined model specification, parameter estimation and interpretation techniques. 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). \mathbf{Y} = \mathbb{E}\left(\mathbf{Y} | \mathbf{X} \right) \mathbb{E} \left[ (Y - \mathbb{E} [Y|\mathbf{X}])^2 \right] = \mathbb{E}\left[ \mathbb{V}{\rm ar} (Y | X) \right]. \mathbf{Y} | \mathbf{X} \sim \mathcal{N} \left(\mathbf{X} \boldsymbol{\beta},\ \sigma^2 \mathbf{I} \right) The prediction interval around yhat can be calculated as follows: 1. yhat +/- z * sigma. &= \mathbb{E}\left[ \mathbb{V}{\rm ar} (Y | X) \right] + \mathbb{E} \left[ (\mathbb{E} [Y|\mathbf{X}] - g(\mathbf{X}))^2\right]. \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}})\\ Thus, $$g(\mathbf{X}) = \mathbb{E} [Y|\mathbf{X}]$$ is the best predictor of $$Y$$. \widehat{Y}_i \pm t_{(1 - \alpha/2, N-2)} \cdot \text{se}(\widetilde{e}_i) A confidence interval gives a range for $$\mathbb{E} (\boldsymbol{Y}|\boldsymbol{X})$$, whereas a prediction interval gives a range for $$\boldsymbol{Y}$$ itself.$, $$\epsilon \sim \mathcal{N}(\mu, \sigma^2)$$, $$\mathbb{E}(\exp(\epsilon)) = \exp(\mu + \sigma^2/2)$$, $$\mathbb{V}{\rm ar}(\epsilon) = \left[ \exp(\sigma^2) - 1 \right] \exp(2 \mu + \sigma^2)$$, $$\exp(0) = 1 \leq \exp(\widehat{\sigma}^2/2)$$. Therefore we can use the properties of the log-normal distribution to derive an alternative corrected prediction of the log-linear model: Home; Uncategorized; statsmodels ols multiple regression; statsmodels ols multiple regression The confidence interval is a range within which our coefficient is likely to fall. \log(Y) = \beta_0 + \beta_1 X + \epsilon We can be 95% confident that total_unemployed‘s coefficient will be within our confidence interval, [-9.185, -7.480]. \begin{aligned} \] pred = results.get_prediction(x_predict) pred_df = pred.summary_frame() Assume that the best predictor of $$Y$$ (a single value), given $$\mathbf{X}$$ is some function $$g(\cdot)$$, which minimizes the expected squared error: Linear regression is a standard tool for analyzing the relationship between two or more variables. $import statsmodels.stats.proportion as smp # e.g. Then, a $$100 \cdot (1 - \alpha)\%$$ prediction interval for $$Y$$ is: \[ regression.$, $$\widetilde{\mathbf{X}} \boldsymbol{\beta}$$, $\[$. Along the way, we’ll discuss a variety of topics, including Then, the $$100 \cdot (1 - \alpha) \%$$ prediction interval can be calculated as: $\[$, $$g(\mathbf{X}) = \mathbb{E} [Y|\mathbf{X}]$$, $\[ 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: &= 0 Fitting and predicting with 3 separate models is somewhat tedious, so we can write a model that wraps the Gradient Boosting Regressors into a single class. \widetilde{\boldsymbol{e}} = \widetilde{\mathbf{Y}} - \widehat{\mathbf{Y}} = \widetilde{\mathbf{X}} \boldsymbol{\beta} + \widetilde{\boldsymbol{\varepsilon}} - \widetilde{\mathbf{X}} \widehat{\boldsymbol{\beta}}$ \begin{aligned} 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 $$X$$. The predict method only returns point predictions (similar to forecast), while the get_prediction method also returns additional results (similar to get_forecast). ... wls_prediction_std calculates standard deviation and confidence interval for prediction. Sorry for posting in this old issue, but I found this when trying to figure out how to get prediction intervals from a linear regression model (statsmodels.regression.linear_model.OLS). or more compactly, $$\left[ \exp\left(\widehat{\log(Y)} \pm t_c \cdot \text{se}(\widetilde{e}_i) \right)\right]$$. \] \end{aligned} Finally, it also depends on the scale of $$X$$. \log(Y) = \beta_0 + \beta_1 X + \epsilon Having obtained the point predictor $$\widehat{Y}$$, we may be further interested in calculating the prediction (or, forecast) intervals of $$\widehat{Y}$$. \] \text{argmin}_{g(\mathbf{X})} \mathbb{E} \left[ (Y - g(\mathbf{X}))^2 \right]. Here is the Python/statsmodels.ols code and below that the results: ... Several models have now a get_prediction method that provide standard errors and confidence interval for predicted mean and prediction intervals for new observations. DONATE &= \mathbb{E}(Y|X)\cdot \exp(\epsilon) Our second model also has an R-squared of 65.76%, but again this doesn’t tell us anything about how precise our prediction interval will be. where: The expected value of the random component is zero. We have examined model specification, parameter estimation and interpretation techniques. Interest Rate 2. Running simple linear Regression first using statsmodel OLS. Y = \beta_0 + \beta_1 X + \epsilon \] \begin{aligned} Collect a sample of data and calculate a prediction interval. Furthermore, since $$\widetilde{\boldsymbol{\varepsilon}}$$ are independent of $$\mathbf{Y}$$, it holds that: # X: X matrix of data to predict. \mathbb{V}{\rm ar}\left( \widetilde{\boldsymbol{e}} \right) &= The get_forecast() function allows the prediction interval to be specified.. \end{aligned} Note that our prediction interval is affected not only by the variance of the true $$\widetilde{\mathbf{Y}}$$ (due to random shocks), but also by the variance of $$\widehat{\mathbf{Y}}$$ (since coefficient estimates, $$\widehat{\boldsymbol{\beta}}$$, are generally imprecise and have a non-zero variance), i.e.Â it combines the uncertainty coming from the parameter estimates and the uncertainty coming from the randomness in a new observation. \widehat{Y}_{c} = \widehat{\mathbb{E}}(Y|X) \cdot \exp(\widehat{\sigma}^2/2) = \widehat{Y}\cdot \exp(\widehat{\sigma}^2/2) ... (OLS - ordinary least squares) is the assumption that the errors follow a normal distribution. The same ideas apply when we examine a log-log model. \widehat{Y} = \exp \left(\widehat{\log(Y)} \right) = \exp \left(\widehat{\beta}_0 + \widehat{\beta}_1 X\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] \\ \widehat{\mathbf{Y}} = \widehat{\mathbb{E}}\left(\widetilde{\mathbf{Y}} | \widetilde{\mathbf{X}} \right)= \widetilde{\mathbf{X}} \widehat{\boldsymbol{\beta}} 35 out of a sample 120 (29.2%) people have a particular… \mathbf{Y} | \mathbf{X} \sim \mathcal{N} \left(\mathbf{X} \boldsymbol{\beta},\ \sigma^2 \mathbf{I} \right) ... from statsmodels. We do … ), government policies (prediction of growth rates for income, inflation, tax revenue, etc.) A first important Assume that the data really are randomly sampled from a Gaussian distribution. &= \sigma^2 \left( \mathbf{I} + \widetilde{\mathbf{X}} \left( \mathbf{X}^\top \mathbf{X}\right)^{-1} \widetilde{\mathbf{X}}^\top\right) Please see the four graphs below. 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. \], \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] 1.96 for a 95% interval) and sigma is the standard deviation of the predicted distribution. \begin{aligned} Y = \exp(\beta_0 + \beta_1 X + \epsilon) statsmodels.regression.linear_model.OLSResults.conf_int ... Returns the confidence interval of the fitted parameters. # Let's calculate the mean resposne (i.e. In practice, you aren't going to hand-code confidence intervals. For larger samples sizes $$\widehat{Y}_{c}$$ is closer to the true mean than $$\widehat{Y}$$. \]. &= \mathbb{C}{\rm ov} (\widetilde{\boldsymbol{\varepsilon}}, \widetilde{\mathbf{X}} \left( \mathbf{X}^\top \mathbf{X}\right)^{-1} \mathbf{X}^\top \mathbf{Y})\\ \], $STAT 141 REGRESSION: CONFIDENCE vs PREDICTION INTERVALS 12/2/04 Inference for coefﬁcients Mean response at x vs. New observation at x Linear Model (or Simple Linear Regression) for the population. Furthermore, this correction assumes that the errors have a normal distribution (i.e.Â that (UR.4) holds). &= \mathbb{E} \left[ (Y - \mathbb{E} [Y|\mathbf{X}])^2 + 2(Y - \mathbb{E} [Y|\mathbf{X}])(\mathbb{E} [Y|\mathbf{X}] - g(\mathbf{X})) + (\mathbb{E} [Y|\mathbf{X}] - g(\mathbf{X}))^2 \right] \\ Overview¶.$, fitted) values again: # Prediction intervals for the predicted Y: #from statsmodels.stats.outliers_influence import summary_table, #dt = summary_table(lm_fit, alpha = 0.05)[1], #yprd_ci_lower, yprd_ci_upper = dt[:, 6:8].T, $$\mathbb{E} (\boldsymbol{Y}|\boldsymbol{X})$$, $$\widehat{\mathbf{Y}} = \mathbb{E} (\boldsymbol{Y}|\boldsymbol{X})$$, $$\widetilde{\mathbf{X}} \widehat{\boldsymbol{\beta}}$$, \[ We can perform regression using the sm.OLS class, where sm is alias for Statsmodels. Because $$\exp(0) = 1 \leq \exp(\widehat{\sigma}^2/2)$$, the corrected predictor will always be larger than the natural predictor: $$\widehat{Y}_c \geq \widehat{Y}$$. We will show that, in general, the conditional expectation is the best predictor of $$\mathbf{Y}$$. We will examine the following exponential model: \[ In our case: There is a slight difference between the corrected and the natural predictor when the variance of the sample, $$Y$$, increases. © Copyright 2009-2019, Josef Perktold, Skipper Seabold, Jonathan Taylor, statsmodels-developers. &= \mathbb{E}\left[ \mathbb{V}{\rm ar} (Y | X) \right] + \mathbb{E} \left[ (\mathbb{E} [Y|\mathbf{X}] - g(\mathbf{X}))^2\right]. Calculate and plot Statsmodels OLS and WLS confidence intervals - ci.py. \end{aligned} \text{argmin}_{g(\mathbf{X})} \mathbb{E} \left[ (Y - g(\mathbf{X}))^2 \right]. Prediction intervals are conceptually related to confidence intervals, but they are not the same. Follow us on FB. 3.7 OLS Prediction and Prediction Intervals, Hence, a prediction interval will be wider than a confidence interval. Prediction intervals must account for both: (i) the uncertainty of the population mean; (ii) the randomness (i.e.Â scatter) of the data., $$\left[ \exp\left(\widehat{\log(Y)} \pm t_c \cdot \text{se}(\widetilde{e}_i) \right)\right]$$, Y &= \exp(\beta_0 + \beta_1 X + \epsilon) \\ Y = \beta_0 + \beta_1 X + \epsilon We can use statsmodels to calculate the confidence interval of the proportion of given ’successes’ from a number of trials. \[ \[ A prediction interval relates to a realization (which has not yet been observed, but will be observed in the future), whereas a confidence interval pertains to a parameter (which is in principle not observable, e.g., the population mean). Let’s use statsmodels’ plot_regress_exog function to help us understand our model. \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] \\ \end{aligned} \[, $$\widehat{\sigma}^2 = \dfrac{1}{N-2} \sum_{i = 1}^N \widehat{\epsilon}_i^2$$, $$\text{se}(\widetilde{e}_i) = \sqrt{\widehat{\mathbb{V}{\rm ar}} (\widetilde{e}_i)}$$, $$\widehat{\mathbb{V}{\rm ar}} (\widetilde{\boldsymbol{e}})$$, $\mathbf{Y} = \mathbb{E}\left(\mathbf{Y} | \mathbf{X} \right) ie., The default alpha = .05 returns a 95% confidence interval. &=\mathbb{E} \left[ \mathbb{E}\left((Y - \mathbb{E} [Y|\mathbf{X}])^2 | \mathbf{X}\right)\right] + \mathbb{E} \left[ 2(\mathbb{E} [Y|\mathbf{X}] - g(\mathbf{X}))\mathbb{E}\left[Y - \mathbb{E} [Y|\mathbf{X}] |\mathbf{X}\right] + \mathbb{E} \left[ (\mathbb{E} [Y|\mathbf{X}] - g(\mathbf{X}))^2 | \mathbf{X}\right] \right] \\$ The difference from the mean response is that when we are talking about the prediction, our regression outcome is composed of two parts: Prediction plays an important role in financial analysis (forecasting sales, revenue, etc. \] # q: Quantile. &= \exp(\beta_0 + \beta_1 X) \cdot \exp(\epsilon)\\ In this lecture, we’ll use the Python package statsmodels to estimate, interpret, and visualize linear regression models.. It’s derived from a Scikit-Learn model, so we use the same syntax for training / prediction… \mathbb{E} \left[ (Y - \mathbb{E} [Y|\mathbf{X}])^2 \right] = \mathbb{E}\left[ \mathbb{V}{\rm ar} (Y | X) \right]. Using the conditional moment properties, we can rewrite $$\mathbb{E} \left[ (Y - g(\mathbf{X}))^2 \right]$$ as: This will provide a normal approximation of the prediction interval (not confidence interval) and works for a vector of quantiles: def ols_quantile(m, X, q): # m: Statsmodels OLS model. Interpreting the Prediction Interval. However, linear regression is very simple and interpretative using the OLS module. sandbox. applies to WLS and OLS, not to general GLS, that is independently but not identically distributed observations Skip to content. In order to do so, we apply the same technique that we did for the point predictor - we estimate the prediction intervals for $$\widehat{\log(Y)}$$ and take their exponent. \begin{aligned} the prediction is comprised of the systematic and the random components, but they are multiplicative, rather than additive. Where yhat is the predicted value, z is the number of standard deviations from the Gaussian distribution (e.g. If you do this many times, youâd expect that next value to lie within that prediction interval in $$95\%$$ of the samples.The key point is that the prediction interval tells you about the distribution of values, not the uncertainty in determining the population mean. from IPython.display import HTML, display import statsmodels.api as sm from statsmodels.formula.api import ols from statsmodels.sandbox.regression.predstd import wls_prediction_std import matplotlib.pyplot as plt import seaborn as sns %matplotlib inline sns.set_style("darkgrid") import pandas as pd import numpy as np [10.83615884 10.70172168 10.47272445 10.18596293 9.88987328 9.63267325 9.45055669 9.35883215 9.34817472 9.38690914] We want to predict the value $$\widetilde{Y}$$, for this given value $$\widetilde{X}$$. 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. &= \mathbb{E}(Y|X)\cdot \exp(\epsilon) From the distribution of the dependent variable: In order to do that we assume that the true DGP process remains the same for $$\widetilde{Y}$$. Let $$\widetilde{X}$$ be a given value of the explanatory variable. \end{aligned} \], $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. Using formulas can make both estimation and prediction a lot easier, We use the I to indicate use of the Identity transform. The sm.OLS method takes two array-like objects a and b as input. Prediction vs Forecasting¶ The results objects also contain two methods that all for both in-sample fitted values and out-of-sample forecasting. \[ In this exercise, we've generated a binomial sample of the number of heads in 50 fair coin flips saved as the heads variable.$, $$\mathbb{E} \left[ (Y - g(\mathbf{X}))^2 \right]$$, $&= \mathbb{C}{\rm ov} (\widetilde{\boldsymbol{\varepsilon}}, \widetilde{\mathbf{X}} \left( \mathbf{X}^\top \mathbf{X}\right)^{-1} \mathbf{X}^\top \mathbf{Y})\\ This is also known as the standard error of the forecast.$, We can defined the forecast error as \widetilde{\mathbf{Y}}= \mathbb{E}\left(\widetilde{\mathbf{Y}} | \widetilde{\mathbf{X}} \right) + \widetilde{\boldsymbol{\varepsilon}} \end{aligned} 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. \] Let our univariate regression be defined by the linear model: ; transform (bool, optional) – If the model was fit via a formula, do you want to pass exog through the formula.Default is True. from statsmodels.sandbox.regression.predstd import wls_prediction_std _, upper, lower = wls_prediction_std (model) plt. Prediction Interval Model. They are predict and get_prediction. &= 0 &= \exp(\beta_0 + \beta_1 X) \cdot \exp(\epsilon)\\ \], $$\mathbb{E}\left(\widetilde{Y} | \widetilde{X} \right) = \beta_0 + \beta_1 \widetilde{X}$$, , $$\mathbb{E}\left[ \mathbb{E}\left(h(Y) | X \right) \right] = \mathbb{E}\left[h(Y)\right]$$, $$\mathbb{V}{\rm ar} ( Y | X ) := \mathbb{E}\left( (Y - \mathbb{E}\left[ Y | X \right])^2| X\right) = \mathbb{E}( Y^2 | X) - \left(\mathbb{E}\left[ Y | X \right]\right)^2$$, $$\mathbb{V}{\rm ar} (\mathbb{E}\left[ Y | X \right]) = \mathbb{E}\left[(\mathbb{E}\left[ Y | X \right])^2\right] - (\mathbb{E}\left[\mathbb{E}\left[ Y | X \right]\right])^2 = \mathbb{E}\left[(\mathbb{E}\left[ Y | X \right])^2\right] - (\mathbb{E}\left[Y\right])^2$$, $$\mathbb{E}\left[ \mathbb{V}{\rm ar} (Y | X) \right] = \mathbb{E}\left[ (Y - \mathbb{E}\left[ Y | X \right])^2 \right] = \mathbb{E}\left[\mathbb{E}\left[ Y^2 | X \right]\right] - \mathbb{E}\left[(\mathbb{E}\left[ Y | X \right])^2\right] = \mathbb{E}\left[ Y^2 \right] - \mathbb{E}\left[(\mathbb{E}\left[ Y | X \right])^2\right]$$, $$\mathbb{V}{\rm ar}(Y) = \mathbb{E}\left[ Y^2 \right] - (\mathbb{E}\left[ Y \right])^2 = \mathbb{V}{\rm ar} (\mathbb{E}\left[ Y | X \right]) + \mathbb{E}\left[ \mathbb{V}{\rm ar} (Y | X) \right]$$, (415) 828-4153 toniskittyrescue@hotmail.com. Assume that the data really are randomly sampled from a Gaussian distribution. In the time series context, prediction intervals are known as forecast intervals. \[ The key point is that the confidence interval tells you about the likely location of the true population parameter. \begin{aligned} \end{aligned} and let assumptions (UR.1)-(UR.4) hold. Adding the third and fourth properties together gives us. \widetilde{\mathbf{Y}}= \mathbb{E}\left(\widetilde{\mathbf{Y}} | \widetilde{\mathbf{X}} \right) + \widetilde{\boldsymbol{\varepsilon}} Parameters: alpha (float, optional) – The alpha level for the confidence interval. statsmodels.sandbox.regression.predstd.wls_prediction_std (res, exog=None, weights=None, alpha=0.05) [source] ¶ calculate standard deviation and confidence interval for prediction. Thanks for reporting this - it is still possible, but the syntax has changed to get_prediction or get_forecast to get the full output object rather than the full_results keyword argument to … $statsmodels logistic regression predict, Simple logistic regression using statsmodels (formula version) Linear regression with the Associated Press # In this piece from the Associated Press , Nicky Forster combines from the US Census Bureau and the CDC to see how life expectancy is related to actors like unemployment, income, and others. 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 $$\widehat{\mathbb{V}{\rm ar}} (\widetilde{\boldsymbol{e}})$$. \[ and so on.$, $\[ Prediction intervals tell you where you can expect to see the next data point sampled. \[ Ie., we do not want any expansion magic from using **2, Now we only have to pass the single variable and we get the transformed right-hand side variables automatically.$ \] Because, if $$\epsilon \sim \mathcal{N}(\mu, \sigma^2)$$, then $$\mathbb{E}(\exp(\epsilon)) = \exp(\mu + \sigma^2/2)$$ and $$\mathbb{V}{\rm ar}(\epsilon) = \left[ \exp(\sigma^2) - 1 \right] \exp(2 \mu + \sigma^2)$$. Linear regression is used as a predictive model that assumes a linear relationship between the dependent variable (which is the variable we are trying to predict/estimate) and the independent variable/s (input variable/s used in the prediction).For example, you may use linear regression to predict the price of the stock market (your dependent variable) based on the following Macroeconomics input variables: 1. Unfortunately, our specification allows us to calculate the prediction of the log of $$Y$$, $$\widehat{\log(Y)}$$. Regression Plots . Let's utilize the statsmodels package to streamline this process and examine some more tendencies of interval estimates.. \mathbb{V}{\rm ar}\left( \widetilde{\mathbf{Y}} - \widehat{\mathbf{Y}} \right) \\ Taking $$g(\mathbf{X}) = \mathbb{E} [Y|\mathbf{X}]$$ minimizes the above equality to the expectation of the conditional variance of $$Y$$ given $$\mathbf{X}$$: