KL-Divergence of Uniform distributions Having $P=Unif[0,\theta_1]$ and  $Q=Unif[0,\theta_2]$ where $0<\theta_1<\theta_2$
I would like to calculate the KL divergence $KL(P,Q)=?$
I know the uniform pdf: $\frac{1}{b-a}$ and that the distribution is continous, therefore I use the general KL divergence formula:
$$KL(P,Q)=\int f_{\theta}(x)*ln(\frac{f_{\theta}(x)}{f_{\theta^*}(x)})$$
$$=\int\frac{1}{\theta_1}*ln(\frac{\frac{1}{\theta_1}}{\frac{1}{\theta_2}})$$
$$=\int\frac{1}{\theta_1}*ln(\frac{\theta_2}{\theta_1})$$
From here on I am not sure how to use the integral to get to the solution.
 A: You got it almost right, but you forgot the indicator functions. So the pdf for each uniform is
$$P(P=x) = \frac{1}{\theta_1}\mathbb I_{[0,\theta_1]}(x)$$
$$\mathbb P(Q=x) = \frac{1}{\theta_2}\mathbb I_{[0,\theta_2]}(x)$$
Hence,
$$
KL(P,Q) = \int_{\mathbb R}\frac{1}{\theta_1}\mathbb I_{[0,\theta_1]}(x)
\ln\left(\frac{\theta_2 \mathbb I_{[0,\theta_1]}}{\theta_1 \mathbb I_{[0,\theta_2]}}\right)dx
$$
Since $\theta_1 < \theta_2$, we can change the integration limits from $\mathbb R$ to $[0,\theta_1]$ and eliminate the indicator functions from the equation. Also, since the distribution is constant, the integral can be trivially solved
$$
\int_{\mathbb R}\frac{1}{\theta_1}\mathbb I_{[0,\theta_1]}
\ln\left(\frac{\theta_2 \mathbb I_{[0,\theta_1]}}{\theta_1 \mathbb I_{[0,\theta_2]}}\right)dx =
\int_{\mathbb [0,\theta_1]}\frac{1}{\theta_1}
\ln\left(\frac{\theta_2}{\theta_1}\right)dx=$$
$$
=\frac {\theta_1}{\theta_1}\ln\left(\frac{\theta_2}{\theta_1}\right) -
\frac {0}{\theta_1}\ln\left(\frac{\theta_2}{\theta_1}\right)=
\ln\left(\frac{\theta_2}{\theta_1}\right)
$$
And you are done.
