Question:
Let XβΌNormal (0, 1) and π = |π|. If the probability density function of π is ππ (y) then for π¦>0,\( \sqrt{\frac{\pi}{2}}\) ππ (π¦) is
Let XβΌNormal (0, 1) and π = |π|. If the probability density function of π is ππ (y) then for π¦>0,\( \sqrt{\frac{\pi}{2}}\) ππ (π¦) is
Updated On: Oct 1, 2024
- \(e^{-\frac{y^2}{2}}\)
- \(e^{\frac{y^2}{2}}\)
- \(e^{-y^2}\)
- \(e^{-\frac{y}{2}}\)
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The Correct Option is A
Solution and Explanation
The correct option is (A): \(e^{-\frac{y^2}{2}}\)
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