Q:

$\log\left(\frac{d\nu}{d\lambda}\right)=-\frac{x^2}{2}$

I have $\log(\frac{d\nu}{d\lambda})$, where $\nu$ is the standard normal distribution and $\lambda$ is the Lebesgue measure. Why is $$\log\left(\frac{d\nu}{d\lambda}\right)=-\frac{x^2}{2}?$$

A:

Note that
$$
d\nu=\exp{(-x^2/2)}\,d\lambda
$$
i.e.
$$
\nu(E)=\int_E \exp{(-x^2/2)}\,d\lambda
$$
for $E$ a Lebesgue measurable set. The measure $\nu$ is induced by the density of the standard normal.

