Related to my most recent post, I have a question I get stuck in.

Let $\phi$ and $\Phi$ be the standard normal density and distribution function, respectively. Show that $f(x)=2\phi(x)\Phi(\lambda x)$ is a probability density function for some random variable ($\lambda,x \in \mathbb{R}$). Hint. Use $\Phi(\lambda x)=1-\Phi (- \lambda x)$ and argue on symmetry.

My attempt was to show that $\int_{-\infty}^{\infty}f(x)dx=1$ for any $x,\lambda$. So, using the hint, $f(x)=2 \phi(x)(1-\Phi(-\lambda x))=2\phi(x)-2\phi(x)\Phi(-\lambda x)$. Then integrating both sides in the range $(-\infty,\infty)$, I get $$\int_{-\infty}^{\infty}2\phi(x)\Phi(\lambda x)dx=\int_{-\infty}^{\infty}\{2\phi(x)-2\phi(x)\Phi(-\lambda x)\}dx$$ which, $$\int_{-\infty}^{\infty}2\phi(x)\Phi(\lambda x)dx=\int_{-\infty}^{\infty}2\phi(x)dx-\int_{-\infty}^{\infty}2\phi(x)\Phi(-\lambda x)dx=2-\int_{-\infty}^{\infty}2\phi(x)\Phi(-\lambda x)dx$$

Here's where I get stuck. If my logic is correct up to here, the only way that my $f(x)$ is a density function is that $\int_{-\infty}^{\infty}2\phi(x)\Phi(-\lambda x)dx=\int_{-\infty}^{\infty}2\phi(x)\Phi(\lambda x)dx$. But clearly $\Phi(\lambda x) \neq \Phi(-\lambda x)$ (unless $\Phi(0)$) since, in the question's hint, we are told that $\Phi(\lambda x)=1-\Phi (- \lambda x)$.

Then that means $\int_{-\infty}^{\infty}f(x)dx \neq 1$ and that $f(x)$ isn't a density function. Which can't be.

I am very confused as I cannot see what and where I have done wrong and why it is wrong. How do I go about solving this problem? Thanks

  • $\begingroup$ $$\int_{-\infty}^{+\infty} f(x)\,dx = \int_0^{+\infty} f(x) + f(-x)\,dx$$ $\endgroup$ Oct 28 '15 at 19:22

It is true that $\Phi(\lambda x) \neq \Phi(-\lambda x)$, but this doesn't prevent $$ \int_{-\infty}^{\infty}2\phi(x)\Phi(-\lambda x)dx=\int_{-\infty}^{\infty}2\phi(x)\Phi(\lambda x)dx\tag1 $$ from being true. To show (1), apply the substitution $u=-x$ to the LHS, and the fact that $\phi(u)$ is symmetric around $u=0$. (i.e., $\phi(-u)=\phi(u)$ for all $u$.)

  • $\begingroup$ Wonderful, thanks!! $\endgroup$
    – Kydo
    Oct 28 '15 at 19:37

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