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!

  • 2
    $\begingroup$ Incorrect: The rhs of the equation is zero since the limit does not depend on $\theta$. $\endgroup$ – Xi'an Feb 2 '16 at 7:36
  • $\begingroup$ Did you perhaps mean to write $\frac{\partial}{\partial a}$ on the right hand side? $\endgroup$ – whuber Feb 2 '16 at 14:28
  • $\begingroup$ math.stackexchange.com/questions/409178 comes very close to answering this question. $\endgroup$ – whuber Feb 2 '16 at 18:00
  • $\begingroup$ @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 $F(x, \theta)$. $\endgroup$ – 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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.