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A log-linear trend model approximated on the interest rate (in %) movement is given as:

\ln I_t = -0.1567 + 0.00134t + \hat{e}_t

for the last 20 years. The standard deviation of the residual is 0.0342. Assuming that the residuals are white noise, what is the point forecast of the interest rate after 3 years from now?

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NNikitesh

Last updated: February 2, 2026 at 10:22

A

0.8589%

B

0.8453%

C

0.7890%

D

0.7945%

Explanation:

The forecast under the log-linear model is given by:

\begin{aligned} \mathbb{E}_T(Y_{T+h}) &= e^{\beta_0 + \beta_1(Y_{T+h}) + \frac{\sigma^2}{2}} \\ \Rightarrow \mathbb{E}_T(I_3) &= e^{\beta_0 + \beta_1(3) + \frac{\sigma^2}{2}} \\ &= e^{-0.1567 + 0.00134(3) + \frac{0.0342^2}{2}} = 0.8589\% \end{aligned}

Explanation:

Model Understanding: This is a log-linear trend model where the natural logarithm of the interest rate is modeled as a linear function of time plus an error term.
Forecast Formula: For log-linear models, when forecasting the original variable (not the log), we need to account for the transformation bias. The correct forecast formula is: $\mathbb{E}_T(I_{T+h}) = e^{\beta_0 + \beta_1(T+h) + \frac{\sigma^2}{2}}$ The term $\frac{\sigma^2}{2}$ is the bias adjustment term needed because $E[e^X] \neq e^{E[X]}$ when X is normally distributed.
Calculation Steps:
- $\beta_0 = -0.1567$
- $\beta_1 = 0.00134$
- $h = 3$ (forecast horizon)
- $\sigma = 0.0342$
- $\sigma^2 = (0.0342)^2 = 0.00116964$
Compute exponent: $-0.1567 + 0.00134 \times 3 + \frac{0.00116964}{2}$ $= -0.1567 + 0.00402 + 0.00058482$ $= -0.15209518$

Compute forecast: $$e^{-0.15209518} = 0.8589 $`$ 4`. Interpretation: The point forecast for the interest rate after 3 years is 0.8589%.

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