Product of cdf and pdf of normal distribution. 
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. 
 A: This question requires no calculations.
You should not integrate anything to answer it.
The Key
If
$\phi(x)$ is the density function of a distribution $D$,
and
$\Phi(x)$ is the cumulative distribution function of $D$,
then
the density $f(x) = 2\phi(x)\Phi(x)$
corresponds to the distribution of a random variable $X$ defined 
as the greater of two random variables (say $D_1,D_2$) picked according to $D$.
Why you would see this
To see this from the formula for $f$, see that $\phi(x)$ corresponds to the chance that $D_1$ has a certain value, and $\Phi(x)$ is the chance that another sample $D_2$ has a lesser value.  Finally, the factor of 2 is for the alternative case that actually $D_2$ had this value and was greater, so together, $f(x)$ is the chance that $x$ is the greater of two randomly picked values.
Once you see this, answering the question is easy and requires no calculations.A: Yes, you expect the higher of two samples to be greater on average than what you expect for just the first sample, since the second sample can only make the maximum go up.  (Remember $0$ is the expected value of a single sample of $N(0,1)$.)
B: No, you don't expect it to be lower.
C: Will the greater of two samples of $N(0,1)$ be negative more than half the time?  Of course not!  (It will be negative 1/4 of the time -- when both samples are negative.)
D: Will the greater sample be positive less than 1/4 of the time?  Of course not!  (It will be positive 3/4 of the time -- when either sample is positive.)
A: An intuitive way to see that A) holds is the following:
Forget about the $2$ in front, it won't change the sign of the integral. Now, if there was only $\phi(x)$, you would get zero, because we know what 
$$\int _\mathbb R x\phi(x) dx = 0$$
Now you multiply the integrand by $\Phi(x)$. Being a cumulative distribution function, it starts from $0$ and goes to $1$. In $0$, it is $0.5$. This means that $\Phi(x)$ is bigger for $x$ positive than for $x$ negative; that is, $x > 0, y < 0 \implies \Phi(x) > \Phi(y)$. Therefore, the positive parts will have bigger weight than the negative parts. And since before they balanced out to give $0$, now the positive part wins and we have 
$$\int _\mathbb R x\phi(x)\Phi(x) dx > 0$$
More formally, 
$$\int _\mathbb R x\phi(x)\Phi(x) dx = \int _{-\infty}^0 x\phi(x)\Phi(x) dx + \int _0^{+\infty} x\phi(x)\Phi(x) dx > 0.5\left(\int _\mathbb R x\phi(x) dx\right) = 0$$
A: Hint:
If $x>0$ then $f(-x)<f(x)$
A: The required expectation is nothing but $2\mathbb E(X\Phi(X))$ where $X\sim\mathcal N(0,1)$.
Integrating by parts ( taking $\Phi(x)$ as 1st function and $x\phi(x)$ as 2nd function) and using the fact that $\phi'(x)=-x\phi(x)$, it can be shown that 
$$\mathbb E(X\Phi(X))=\int_{\mathbb R}x\Phi(x)\phi(x)\,\mathrm{d}x$$
$$\qquad\qquad\qquad\quad=\int_{\mathbb R}\frac{1}{2\pi}e^{-x^2}\,\mathrm{d}x=\frac{1}{2\sqrt{\pi}}$$ 
Edit.
Integrating by parts, $\displaystyle\int_{\mathbb{R}}\Phi(x)x\phi(x)\,\mathrm{d}x=\lim_{A\to\infty}\left(-\Phi(x)\phi(x)\big|_{-A}^A\right)+\int_{\mathbb{R}}\phi(x)\phi(x)\,\mathrm{d}x$
The limit is $0$ because $\phi(x)\to0$ whenever $x\uparrow\infty$ or $x\downarrow-\infty$
A: 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:
$$\begin{align*} E[X] &= \int_0^\infty (1-\Phi^2(x)) dx - \int_{-\infty}^0 \Phi^2(x) dx \\ &= \int_0^\infty (1-\Phi^2(x)) dx - \int_{0}^\infty \Phi^2(-x) dx \\ &= \int_0^\infty (1-\Phi^2(x)) dx - \int_{0}^\infty (1-\Phi(x))^2 dx \\ &= 2 \int_0^\infty \Phi(x)(1-\Phi(x)) dx  > 0\end{align*}$$
