Estimating a triangular half-width from the spread

Question

You model a symmetric shock as a triangular distribution on $(-\theta, \theta)$ that peaks at $0$ , with unknown half-width $\theta$ . Its variance is $\theta^2 / 6$ . From your data the average of the squares is $\frac{1}{n}\sum X_i^2 = 6$ .

Use the method of moments to estimate $\theta$ .

Accepted Answer

The mean tells you nothing. The distribution is symmetric about 0, so its mean is 0 for every 	heta. The first moment equation is empty; use the second moment. Match the second moment. With mean 0, the second moment equals the variance, which for this triangular shape is 	heta^2 / 6: \mathbb{E}[X^2] = \frac{	heta^2}{6}. Set that equal to the sample average of the squares m2 = 6: \frac{	heta^2}{6} = 6 \implies 	heta^2 = 36 \implies \hat	heta = 6. Check. A symmetric triangular on (-6, 6) has variance \frac{6^2}{6} = 6, matching the data. Contrast with the uniform. A uniform on (-	heta, \…

Estimating a triangular half-width from the spread

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