- CFA Exams
- 2024 Level I
- Topic 1. Quantitative Methods
- Learning Module 10. Simple Linear Regression
- Subject 5. Prediction Using Simple Linear Regression and Prediction Intervals

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##### Subject 5. Prediction Using Simple Linear Regression and Prediction Intervals PDF Download

Suppose we have the following regression equation for the Ivory Tower Mutual Fund.

Y

_{t}= 0.0009 + 0.6154 x X

_{(1,t)}+ 0.3976 x X

_{(2,t)}

Where:

X

_{(1,t)}= return on the small-cap value index

X

_{(2,t)}= return on the small-cap growth index

Suppose further that we know that in a given month, the small-cap value index returned -1.2% and the small-cap growth index returned -1.9%. What is the predicted return for the Ivory Tower Mutual Fund?

The return is simply 0.0009 + 0.6154 (-0.012) + .3976 (-0.019) = -1.4%.

Note that although the prediction of an individual value Y

_{i}and the estimate of the mean value E(Y

_{i}| X

_{i}) are the same for a given value X

_{i}, the sampling errors associated with the two predictions are different! The estimate of the mean value of Y

_{i}does not require an estimate of the random error e

_{i}. The confidence interval for the individual Y is always wider than that for mean value.

We often wish to have a confidence interval for the dependent variable for which we are performing the regression. Obtaining a confidence interval for the dependent variable is a bit more complicated because there are multiple sources of variation. First, there is the variation from the standard error of estimate (SEE) in the regression equation. A second source of uncertainty comes from the variation in the estimate of the regression parameters themselves. If these were known with certainty, then the predicted values of the dependent variable would have variance equal to the squared standard error of estimate. However, the parameters themselves are not known with certainty.

The formula for the variance in the prediction error, (s

_{f})

^{2}, of Y, is:

(s

_{f})

^{2}= s

^{2}x {1 + 1/n + [(X - X-bar)

^{2}] / [(n - 1) x (σ

_{x})

^{2}]}

Where:

s

^{2}= squared standard error of estimate

n = number of observations

X = value of the independent variable used for the specific prediction of Y

X-bar = mean of the independent variable across the observations

(σ

_{x})

^{2}= sample variance of the independent variable

Once the variance in the prediction error, (s

_{f})

^{2}, is known, the confidence interval for the dependent variable Y is constructed in a very similar way to the construction of confidence intervals around parameters. The confidence interval will be: Y-hat +or- (t

_{c}) x (s

_{f})

Where Y-hat is the predicted value of Y, based on a particular value of X, t

_{c}is the critical value for the t-statistic for a significance level of alpha, and s

_{f}is the standard error of the predicted value.

*Example*

Suppose we are predicting the excess return on GE stock for the next month using the formula: R

_{GE}- R

_{f}= α

_{GE}+ β

_{GE}x (R

_{m}- R

_{f}) + error

You are given the following data regarding 24 monthly observations of the excess return on GE stock and on the S&P 500 over the risk-free rate:

- The mean excess return on the S&P 500 during the observation period was -0.6751%.
- The variance of the excess return on the S&P 500 during the observation period was 0.2453%.
- The standard error of estimate is 0.0622.
- You expect the excess return on the S&P 500 to be 0.1580% next month.
- The α for GE stock is 1.68%; the β for GE stock is 1.3487.

Find the 95% confidence interval for the excess return on GE stock next month.

*Solution*

First, we predict the value of the dependent variable for next month: Y-hat = R

_{GE}- R

_{f}, the excess return on GE stock. Using the formula and data above we have: 0.0168 + 1.3487(.00158) = 1.893%

Second, compute the variance of the error in this prediction.

(s

_{f})

^{2}= s

^{2}x {1 + 1/n + [(X - X-bar)

^{2}] / [(n-1)x (s

_{x})

^{2}]} = 0.0622

^{2}x {1 + 1/24 + [(0.00158-(-0.006751)]

^{2}/ [(24 - 1) x (0.002453)]} = 0.42483%

The standard deviation of this forecast error is the square root of this number, or 6.5179%.

Third, find the critical value for the t-statistic. We will have 22 degrees of freedom, and the value for a 95% confidence interval will be 2.074.

Fourth, calculate the prediction interval. It will be: 1.893% +or- 2.074 x (6.5179%) = [-11.63%, 15.41%]

This is a fairly large interval for just one month of excess return. Clearly, the large standard deviation in forecast error is a big factor in the size of the interval. It is also due in part to having only 24 observations. By collecting a greater number of observations, we could probably lower the standard deviation of forecast error (and, slightly, the critical value for t).

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**User Contributed Comments**
1

User |
Comment |
---|---|

MaksimGul |
Has anyone got an answer 0.42483%? I got 0.00403480103 or 0.40348% |

I passed! I did not get a chance to tell you before the exam - but your site was excellent. I will definitely take it next year for Level II.