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If we define a random variable $X \sim N(0,1)$ with $\Phi$ being the cdf of a standard normal, what would $E(\Phi(a+bX))$ be?

I was only able to rewrite $\Phi(a+bX)$ as $P(Z\leq aX+b|X)$ with $Z\sim N(0,1)$.

I could also calculate the cdf of $\Phi(a+bX)$ as per this answer but it doesn't help with its expectation.

Any help would be greatly appreciated.

Noah Bishop
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    See https://math.stackexchange.com/questions/449875/expected-value-of-normal-cdf/1125935, https://stats.stackexchange.com/questions/61080/how-can-i-calculate-int-infty-infty-phi-left-fracw-ab-right-phiw. The answer is $\Phi(\frac{a}{\sqrt{1+b^2}})$. – StubbornAtom Jan 22 '20 at 14:20

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For $b\ne 0$,

\begin{align} \mathsf{E}\Phi(a+bX)&=\int_0^{\infty}\mathsf{P}(\Phi(a+bX)>t)\,dt \\ &=\int_0^1\mathsf{P}(X>(\Phi^{-1}(t)-a)/b)\,dt \\ &=1-\int_0^1\Phi((\Phi^{-1}(t)-a)/b)\,dt. \end{align}

When $a=0$ and $b=1$, the third line reduces to $$ 1-\int_0^1 t\,dt=\frac{1}{2}. $$