Integral calculations with exponentials ($\int_{-\infty}^{+\infty}e^{ax^2+bx+c-1}$)

Question

I have a probability distribution function (pdf) $f(x)$ wich has the following form:

$$f(x)=e^{ax^2+bx+c-1}$$

Let's assume that the pdf has a mean value $\mu$ and a variance value $\sigma^2$.

I want to prove that this pdf is the Gaussian function. So, I want to show that:

$$f(x)=\frac{1}{\sqrt{2\pi}\sigma}exp\left \{\frac{-(x-\mu)^2}{2\sigma^2}\right \}$$

I consider finding the parameters $a$, $b$ and $c$ by making use of the following equations: $$\int_{-\infty}^{+\infty}f(x)=1$$ $$\int_{-\infty}^{+\infty}xf(x)=\mu$$ $$\int_{-\infty}^{+\infty}x^2f(x)=\sigma^2 + \mu^2$$

However, I find diffuculty on calculating these integrals since I am not very familiar with such exponential integrations that have a polynomial exponent with a 2nd or higher degree.

For instance, how do I have to approach the first one ($\int_{-\infty}^{+\infty}f(x)=1$)? Do I have to make use of the Gaussian integral somehow? ($\int_{-\infty}^{+\infty}e^{-x^2}=\sqrt{\pi}$). It seems to me that the integral is divergent.

For the other two equations do I have to make use of integration by parts and if so how do I have to approach it?

Any help is appreciated!Thanks in advance!

Answer 1

First, note that $a<0$ or else the integral would diverge. Instead of treating the three equations separately, we can combine them to get some easier integrals, Let

$$A=\int_{-\infty}^\infty f(x)\ dx,\quad B= \int_{-\infty}^\infty xf(x)\ dx,\quad C=\int_{-\infty}^\infty x^2 f(x)\ dx$$

Notice that $2a B+b A=0$ because

$$\int_{-\infty}^\infty (2ax+b)e^{a x^2+bx +c-1}\ dx = \bigg [e^{a x^2+bx +c-1}\bigg]_{-\infty}^{\infty}=0$$

Which gives you the necessary condition $2a\mu+b=0$ or $a=\frac{-b}{2\mu}$ (This means that $b>0$). Now we can write $f$ as

$$f(x)= \exp\left(-\frac{b}{2\mu}\left(x^2-2\mu x+\mu^2\right)+c-1+\frac{b\mu}{2}\right)= d \exp\left(-\frac{b}{2\mu}\left(x-\mu\right)^2\right)$$

(I have replaced $e^{c-1+\frac{b\mu}{2}}$ by d). You can use $A$ to find that $d= \frac{1}{\sqrt{2\pi}} \sqrt{\frac{b}{\mu}}$. Now the last piece of the puzzle is $b$. Notice that

$$\sigma^2 =\int_{-\infty}^{\infty} d(x^2-\mu x) \exp\left(-\frac{b}{2\mu}\left(x-\mu\right)^2\right)\ dx = d \left(\frac{\mu}{b}\right)^{3/2} \sqrt{2\pi}=\frac{\mu}{b}$$

Now, $b=\frac{\mu}{\sigma^2}$, and your desired resuult follows immediately from this.

Answer 2

$$f(x)=ax^2+bx+c+1=\left(\sqrt{a}x+\frac{b}{2\sqrt{a}}\right)^2+\left(c+1-\frac{b^2}{4a}\right)$$ now if we have: $$\int_{-\infty}^\infty e^{f(x)}dx$$ we can use the substitution: $$u=\sqrt{a}x+\frac{b}{2\sqrt{a}}\Rightarrow du=\sqrt{a}dx\Rightarrow dx=\frac{du}{\sqrt{a}}$$ and our integral becomes: $$\int_{-\infty}^\infty\frac{1}{\sqrt{a}}e^{u^2+(c+1-b^2/4a)}du=\frac{\exp\left(c+1-\frac{b^2}{4a}\right)}{\sqrt{a}}\int_{-\infty}^\infty e^{u^2}du$$ now this last integral is definately divergent and is easy to show, I have however laid it out like this incase you meant for the polynomial to be negative which would lead you to the gaussian integral :)