# 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.

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*}

• B is the correct answer? – Ronald Jan 11 '18 at 10:14
• $$E[X] = \int_{-\infty}^\infty x \cdot \Phi'(x)^{2}\ dx =$$ Now by parts $$0 - \int_{-\infty}^\infty 1 \cdot \Phi(x)^{2}\ dx < 0$$ ? right? – Ronald Jan 11 '18 at 10:25
• No, with integration by parts you won't get $$0 - \int_{-\infty}^\infty 1 \cdot \Phi(x)^{2}\ dx < 0$$ but $$\infty - \int_{-\infty}^\infty 1 \cdot \Phi(x)^{2}\ dx < 0 = \infty - \infty$$ what is undefined. I edited my answer a bit to be correct… the first inequality was just wrong. – Gono Jan 11 '18 at 10:48
• $$E[X] = \int_0^\infty (1-F(x)) dx$$ just hold for non-negative random variables. In general for random variable with densities $f$ it holds $$E[X] = \int_0^\infty (1-F(x)) dx - \int_0^\infty F(-x) dx = \int_0^\infty (1-F(x)) dx - \int_{-\infty}^0 F(x) dx$$ what easily can be seen by using Fubinis Theorem and $$F(x) = \int_{-\infty}^x f(x) dx$$ – Gono Jan 11 '18 at 11:18
• We had $$P(X \le x) = \Phi^2(x)$$ just plug in $x=0$ to get: $$P(X \le 0) = \Phi^2(0) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ – Gono Jan 29 '18 at 17:27

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.)

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,

$$\int _\mathbb R x\phi(x)\Phi(x) dx = \int _{-\infty}^0 x\phi(x)\Phi(x) dx + \int _0^{+\infty} x\phi(x)\Phi(x) dx > 0.5\left(\int _\mathbb R x\phi(x) dx\right) = 0$$

• That was very helpful thanks i am new to these things of statistics and in simple language you wrote thanks again. – Ronald Jan 11 '18 at 14:48
• @Daman You're welcome :-) – Ant Jan 11 '18 at 14:50

Hint:

If $x>0$ then $f(-x)<f(x)$

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 $x\downarrow-\infty$

• $\Phi'(x)\not=-x\phi(x)$ – NCh Jan 11 '18 at 11:48
• @NCh Definitely not. I meant $\phi$ in the l.h.s. – StubbornAtom Jan 11 '18 at 11:52
• Can you show how your second integral is deduced from the first one? – NCh Jan 11 '18 at 11:59
• @NCh I added a couple of lines. – StubbornAtom Jan 11 '18 at 12:09
• Thank you, very elegant. I did not know that the expectation of the maximum of two independent normal can be found analytically. – NCh Jan 11 '18 at 12:21