6. consider neyman-pearson criteria for two cauchy distributions in one dimension assume a zero-one error loss, and for simplici
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Answer: choice A
f(x) = (1/4)x^2, f(x) = -0.5x^2, f(x) = x^2
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The coefficient in front of the x^2 term tells us how wide or narrow the parabola will be. It turns out that the further the coefficient value is from 0, in either direction, this will make the parabola more narrow.
Notice how 1/4 = 0.25 which is less than |-0.5| = 0.5 meaning that f(x) = (1/4)x^2 is wider than f(x) = -0.5x^2
Then comparing f(x) = -0.5x^2 to f(x) = 1x^2, we see that 0.5 is smaller than 1, so f(x) = -0.5x^2 is wider than f(x) = 1x^2
That is why the order from widest to narrowest is f(x) = (1/4)x^2, f(x) = -0.5x^2, f(x) = x^2
A graph shown below helps confirm the answer (note the color coding). In figure 1 is the graph of all three functions together. In figure 2, I reflect the second function help show the comparison of why it's the middle-most in terms of how wide it is (it is the orange parabola, which is the reflection of the red parabola)
