Assume $p(x)$ is a probability density function (pdf) with support $[0,\infty)$. How can we show that

$$\int\limits_{x=0}^{\infty}\bigg|p(x)-(1-a)q\left({x}\right)-{a}({1-a}) (q*q)\left({x}\right)\bigg| = \mathcal{O}(a^2)$$ as $a \to 0$, where $*$ denotes convolution, and $q(x)=\frac{p(\frac{x}{1-a})}{1-a}$ is the scaled version of $p(x)$, and $(q*q)(x)$ is the pdf of sum of two independent random variables with pdf $q(x)$.

Assumptions: $p(x)$ is such that if it is scaled to $p_1(x,a)=(1-a) p\left(\left(1-a\right)x\right)$, where $0<a<1$, then \begin{align} \label{c1}\text{• }&\frac{\partial \log{p_1}}{ \partial a}, \frac{\partial^2 \log{p_1}}{\partial a^2}, \frac{\partial^3 \log{p_1}}{\partial a^3} \text{ exist, } \forall a\in (0,1)\\ \text{• }\nonumber & \forall a \in (0,1), \bigg|\frac{\partial p_1}{\partial a}\bigg| < F(x), \text{ s.t. } \int_{x=0}^{\infty}F(x)dx<\infty, \\ &\nonumber \bigg|\frac{\partial^2 p_1}{\partial a^2}\bigg| < G(x),\text{ s.t. } \int_{x=0}^{\infty}G(x)dx<\infty \\ &\nonumber \bigg|\frac{\partial^3 \log p_1}{\partial a^3}\bigg| < H(x), \text{ s.t. } \int_{x=0}^{\infty}p(x)H(x)dx<M<\infty\\ \label{c2} & \text{ where }M \text{ is independent of }a\\ \label{c3}\text{• } & \int_{x=0}^{\infty}\frac{\partial p_1(x,a)}{\partial a}\bigg|_{a=0}dx= \int_{x=0}^{\infty}\frac{\partial^2 p_1(x,a)}{\partial a^2}\bigg|_{a=0}dx=0 \end{align}

I have tried taking derivative of the term with respect to $a$ and evaluating that at $a=0$, but didn't give me the answer.

  • $\begingroup$ Your notation is nonstandard, so please explain what you mean by "convolution". Normally a convolution of two functions is another function, which therefore would have a single argument, but your convolution operator involves two functions that have already been evaluated. Your uses of $p$ also suggest it might be a density function rather than a distribution function. Could you clarify that? $\endgroup$ – whuber Sep 7 '16 at 22:56
  • $\begingroup$ @whuber sure, I edited the question. $\endgroup$ – Sus20200 Sep 8 '16 at 16:28
  • 1
    $\begingroup$ Thank you. It's still not clear what you mean by "$q\left({x}\right)*q\left({x}\right)$." Is that intended to be $(q*q)(x)$? Or is it intended to be the distribution function of the sum of two independent variables (with density $p$) evaluated at $x/(1-a)$? (The two are different.) Perhaps if you were to explain--in words--what this integral represents, there wouldn't be so many difficulties with the notation. $\endgroup$ – whuber Sep 8 '16 at 17:39
  • $\begingroup$ @whuber Yes, by q(x)*q(x), I meant $(q*q)(x)$. It is an alternative notation. But, I edited the problem statement again. $\endgroup$ – Sus20200 Sep 9 '16 at 13:17
  • $\begingroup$ @whuber I also meant that $p(x)$ is density function, not distribution function. $\endgroup$ – Sus20200 Sep 9 '16 at 13:30

Your Answer

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

Browse other questions tagged or ask your own question.