When $\psi(\lambda) = \int_{-\infty}^{\infty}\{\Phi(\lambda x)\}^{K}\phi(x)dx$ is maximized or minimized with respect to $\lambda$? Where $\Phi()$ and $\phi()$ are usual distribution function and density function of standard normal distribution.
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Do you have a good reason why this should have a closed form solution? – Ian Aug 03 '17 at 14:21
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It may not have a closed form. I am not sure. – Satya Prakash Aug 03 '17 at 14:45
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I mean, if it does have a nontrivial closed form solution, it would be obtained by just taking the derivative and setting it equal to zero. But that means solving $\int_{-\infty}^\infty x K \phi (\lambda x)^{K-1} \phi(x) dx=0$ for $\lambda$. Why should that be analytically tractable? – Ian Aug 03 '17 at 14:49
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Some observations: at $\lambda=\pm \infty$ you have $1/2$. At $\lambda=0$ you have $(1/2)^K$. Is it possible that the function is monotone in between? – Ian Aug 03 '17 at 14:59
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1For K = 1, we have $\psi (\lambda) = 1 / 2$ – Kroki Aug 03 '17 at 15:03
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How did you get this Youem?? – Satya Prakash Aug 03 '17 at 16:39
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@SatyaPrakash I add an aswer – Kroki Aug 03 '17 at 16:54
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Using the parity of $\phi$, we have :
$\psi (\lambda) = \int_{0}^{+\infty}(\{\Phi(\lambda x)\}^K + \{\Phi(-\lambda x)\}^K) \phi (x) dx $. Then $\psi$ is an even function it is enough to study the function on $[0,+\infty)$
$\mbox{using } \Phi(-x) = 1 - \Phi (x)$, we have $\psi(\lambda) = \int_{0}^{+\infty} (\{\Phi(\lambda x)\}^K + \{1 - \Phi(\lambda x)\}^K)\phi(x) dx$.
Let $g(u) = u^K +(1-u)^K$, then $g'(u) = K (u^{K-1} - (1-u)^{K-1}) > 0$ for all $u\in (\frac{1}{2}, 1)$ and so $\frac{1}{2 ^ {K-1}} = g(\frac{1}{2}) < g(u) < g(1) = 1$. And so $\frac{1}{2^K} = \frac{1}{2 ^ {K-1}} \int_{0}^{+\infty} \phi(x) dx \le \psi (\lambda) \le \int_{0}^{+\infty} \phi(x) dx = \frac{1}{2}$. This prove that the minimum is $\frac{1}{2^K}$ for $\lambda = 0$ and the maximum $\frac{1}{2}$ for $\lambda = +\infty$
Kroki
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Hi Youem, If we can break $\psi(\lambda)$ in this way then I think it is constant with respect to $\lambda$ for finite and not equal to 0. – Satya Prakash Aug 06 '17 at 14:06
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@SatyaPrakash I am not sure that I have understood your comment. Could you please clarify ? – Kroki Aug 06 '17 at 14:43
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If $0<\lambda<\infty$, then I think $\psi(\lambda)$ is not changing wrt to $\lambda$. – Satya Prakash Aug 06 '17 at 15:10
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I am not very sure about that because the value $\psi (0) = \frac{1}{2^K}$ and the limit to $\infty$ is $\frac{1}{2}$ and $\psi$ is continuous. – Kroki Aug 06 '17 at 15:12