# Question

With reference to Example 9.7, show that if the losses are proportional to the squared errors instead of their absolute values, the risk function becomes

And its minimum is at k = 3/2.

Example 9.7

A random variable has the uniform density

And we want to estimate the parameter θ (the “move” of Nature) on the basis of a single observation. If the decision function is to be of the form d(x) = kx, where k ≥ 1, and the losses are proportional to the absolute value of the errors, that is,

L(kx,θ) = c|kx – θ|

Where c is a positive constant, find the value of k that will minimize the risk.

And its minimum is at k = 3/2.

Example 9.7

A random variable has the uniform density

And we want to estimate the parameter θ (the “move” of Nature) on the basis of a single observation. If the decision function is to be of the form d(x) = kx, where k ≥ 1, and the losses are proportional to the absolute value of the errors, that is,

L(kx,θ) = c|kx – θ|

Where c is a positive constant, find the value of k that will minimize the risk.

## Answer to relevant Questions

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