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Let $X$ be a standard normal random variable and $c$ be a real number. Show that $$\mathbb E((X-c)_+) = g(c) := \frac{1}{\sqrt{2\pi}} e^{-c^2/2}-cN(-c).$$

I am curious though because I thought if $c = 0$, it implies that expectation should be $$ \Bbb E(X_+) = \frac{\sqrt2}{\sqrt\pi} $$

And that doesn't seem to be the result.

Em.
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quaz55
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  • $\mathbb{E}[X_+]=\frac{1}{\sqrt{2\pi}}$. Perhaps you are confusing it with $\mathbb{E}[|X|]$. – carmichael561 Nov 14 '16 at 04:09
  • Yes, but isn't X+ the same as |X|? The link shows I feel it should be different.http://math.stackexchange.com/questions/88114/expected-value-of-normal-distribution-given-that-distribution-is-positive – quaz55 Nov 14 '16 at 04:33
  • The question in your link is computing a conditional expectation, which explains the extra factor of $2$. – carmichael561 Nov 14 '16 at 04:35
  • But isn't this problem a conditional expectation? – quaz55 Nov 14 '16 at 04:36
  • No, it's the expectation of $X_+$. It is related to the conditional expectation by $\mathbb{E}[X_+]=\mathbb{E}[X\mid X\geq 0]\mathbb{P}(X\geq0)$. – carmichael561 Nov 14 '16 at 04:40

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