Property of the expectation of a pdf function

Question

I'm facing a equation look as the follows $$ G(N)=\int_x F(x)^{N-2} f^2(x)dx $$ $F(x)$ is a cdf function and $f$ is its pdf.

I'm curious about how the term changes with $N$. For instance, when $x$ follows uniform distribution, $G(N)$ is proportional to $1/(J-1)$. This example is easy as $f$ is just a constant.

But in general, how the term will be changed with $N$ if we embed a $f$ in the integral? The term $f^2$ makes me confused. Looks like I'm working with $$ \mathbb{E}[F(x)^N\cdot f(x)] $$

Answer 1

$$H(N) = \int_{-\infty}^{\infty} F(x)^{N-2}f(x)dx = \frac{F(x)^{N-1}}{N-1}\bigg\vert_{-\infty}^{\infty} = \frac{1}{N-1} $$

$$G(N) = \int_{-\infty}^{\infty}F(x)^{N-2}f(x)^2dx$$

$$\lim_{N\rightarrow \infty} G(N) = \lim_{N\rightarrow \infty}\int_{-\infty}^{\infty}F(x)^{N-2}f(x)^2dx = \int_{-\infty}^{\infty}\lim_{N\rightarrow \infty}F(x)^{N-2}f(x)^2dx = 0$$

$$\frac{\partial^{r} G(N)}{\partial N^{r}} = \int_{-\infty}^{\infty}\frac{\partial^{r} F(x)^{N-2}f(x)^2}{\partial N^{r}}dx = \int_{-\infty}^{\infty}F(x)^{N-2}f(x)^{2}\log(F(x))^{r}dx$$

Above two observations indicate that,

• As $N \rightarrow \infty$, $G(N) \rightarrow 0$ which can also be seen intutively - As $N$ tends to infinity $F(x) \in [0,1]$ will tend to $0$ for $F(x) \in [0,1)$ making the overall integral to be $0$, and when $F(x)=1$ which will either happen when $x \geq a$ in which case $f(x) = 0$ for $x>a$ (so the area under curve will be $0$) or will happen as $x \rightarrow \infty$ in which case $f(x) \rightarrow 0$ making the overall limit to approach $0$, again making the integral to be $0$.

• The function $G(N)$ is infinitely times differentiable with respect to $N$ and taking derivative odd number of times wrt $N$ gives negative $G(N)$ and even number of times gives positive $G(N)$.

Now, I think the exact value of $G(N)$ depends on the distribution but $G(N)$ is inversely proportional to $N$ and I have this feeling that the exact form of $G(N)$ will either be of type $e^{-N}$ or $\frac{1}{N}$ because both approach $0$ as $N \rightarrow \infty$, both are infinitely differentiable and taking derivative wrt $N$ odd number of times lead to negative $G(N)$ and even number of times lead to positive $G(N)$.

Here, I compute $G(N)$ for some popular distributions,

• $\mathcal{N}(0,1)$ : In the notation below $f(z) = \frac{1}{\sqrt{2\pi}}e^{-z^2/2}$ and $g(z) = f(z)^2 = \frac{1}{2\sqrt{\pi}\sqrt{2\pi}\frac{1}{\sqrt{2}}}e^{-z^2}$ which implies that $g(z)$ is the pdf of $\mathcal{N}(0, \frac{1}{2})$. Let $X, Y_1, Y_2, \ldots, Y_{N-2} \sim \mathcal{N}(0,1)$, $U \sim \mathcal{N}(0, \frac{1}{2})$, $Y_1, Y_2, \ldots, Y_{N-2}, X$ be independent and $Y_1, Y_2, \ldots, Y_{N-2}, U$ be independent.

$$\begin{align} G(N) &= \int_{-\infty}^{\infty}\int_{-\infty}^{x}\int_{-\infty}^{x}\ldots \int_{-\infty}^{x} f(y_1)f(y_2)\ldots f(y_{N-2})f(x)^2dy_1dy_2\ldots dy_{N-2}dx\\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{u}\int_{-\infty}^{u}\ldots \int_{-\infty}^{u} f(y_1)f(y_2)\ldots f(y_{N-2})g(u)dy_1dy_2\ldots dy_{N-2}du \\ &= P(Y_1-U\leq 0, Y_2-U \leq 0, \ldots, Y_{N-2}-U\leq 0) \\ &= P(Z_1\leq0, Z_2\leq 0, \ldots, Z_{N-2}\leq 0) \\ &= \Phi_{Z_1, Z_2, \ldots, Z_{N-2}}(0)\end{align}$$

where $Z_1, Z_2, \ldots Z_{N-2} \sim \mathcal{N}(0,\Sigma)$ where,

$$\Sigma_{ij} = \left\{\begin{matrix}3/2 & i=j \\ 1/2 & i\neq j\end{matrix}\right.$$

Following is a plot of $G(N)$ vs $N$ when $X \sim \mathcal{N}(0,1)$ that I obtained through simulation,

• $\exp(1)$

$$G(N) = \int_{0}^{\infty}(1-e^{-x})^{N-2}e^{-2x}dx = \int_{0}^{1}t^{N-2}(1-t)dt = \frac{1}{N(N-1)}$$

Following is a plot of $G(N)$ vs $N$ when $X \sim \exp{1}$ that I obtained through simulation,

Let me know if you want me to simulate with some continuous distribution other than exponential family.