Probability – Product of CDF and PDF of Normal Distribution

Question

A continuos random variable $X$ has the density
$$
f(x) = 2\phi(x)\Phi(x), ~x\in\mathbb{R}
$$
then

(A) $E(X) > 0$

(B) $E(X) < 0$

(C) $P(X\leq 0) > 0.5$

(D) $P(X\ge0) < 0.25$

\begin{eqnarray}
\Phi(x) &=& \text{Cumulative distribution function of } N(0,1)\\
\phi(x) &=& \text{Density function of } N(0, 1)
\end{eqnarray}

I don't have a slightest clue where to start with. Can someone give me a little push. I saw some answers on same question like this but I didn't understand how should I integrate it when calculating expectation.

Answer 1

It holds $$\left(\Phi^2(x)\right)' = 2\Phi(x)\varphi(x)$$

So we have $$P(X \le x) = \Phi^2(x)$$

And we can conclude $$P(X \le 0) = 0.25$$ $$P(X \ge 0) = 1 - P(X < 0) = 0.75$$

For the expectation we get:

$$\begin{align*} E[X] &= \int_0^\infty (1-\Phi^2(x)) dx - \int_{-\infty}^0 \Phi^2(x) dx \\ &= \int_0^\infty (1-\Phi^2(x)) dx - \int_{0}^\infty \Phi^2(-x) dx \\ &= \int_0^\infty (1-\Phi^2(x)) dx - \int_{0}^\infty (1-\Phi(x))^2 dx \\ &= 2 \int_0^\infty \Phi(x)(1-\Phi(x)) dx > 0\end{align*}$$