# Interchanging limit and derivative for CDFs

Let $F_{\theta}(x)$ denote a cumulative distribution function indexed by the parameter vector $\theta$. Given this definition is the following equation correct (and if so under which conditions)?

$$\lim_{\theta \rightarrow \theta^*}\frac{\partial}{\partial \theta} F_{\theta}(x) \stackrel{???}{=} \frac{\partial}{\partial \theta^*} \lim_{\theta \rightarrow \theta^*} F_{\theta}(x)$$

Thank you very much!

• Incorrect: The rhs of the equation is zero since the limit does not depend on $\theta$. – Xi'an Feb 2 '16 at 7:36
• Did you perhaps mean to write $\frac{\partial}{\partial a}$ on the right hand side? – whuber Feb 2 '16 at 14:28
• math.stackexchange.com/questions/409178 comes very close to answering this question. – whuber Feb 2 '16 at 18:00
• @whuber I thought so too at first, and even grabbed Apostl to help type an answer. Upon further thought, I think it's more simple than that, it's really just about the two variable function $F(x, \theta)$. – Matthew Drury Feb 2 '16 at 18:28

I believe you can exchange the limit and the derivative as long as your function is continuously differentiable in $\theta$ at $\theta^*$. That is, both your CDF and its corresponding PDF need to be continuous at whatever value of $\theta^*$ you are looking at. Assuming this is true, you will have: $$\lim_{\theta \rightarrow \theta^*}\frac{\partial}{\partial \theta} F_{\theta}(x) {=}\frac{\partial}{\partial \theta^*} F_{\theta^*}(x)$$ and $$\frac{\partial}{\partial \theta^*} \lim_{\theta \rightarrow \theta^*} F_{\theta}(x){=}\frac{\partial}{\partial \theta^*} F_{\theta^*}(x)$$ because by definition, $\lim_{x \rightarrow a}f(x){=}f(a)$ for a continuous function.
Note that whether or not the function is continuous in $x$ doesn't matter. For example, the $poisson(\lambda)$, which only has a single continuous parameter $\lambda$, would be one you could exchange the limit on for any valid value of $\lambda$ even though it is not continuous in $x$, but the $\chi^2(k)$ would not be interchangeable everywhere despite being a continuous distribution in $x$ since $k$ is a discrete, discontinuous parameter.