Probability density function: Normal distribution

Question:
When a stochastic variable, $X$, obeys the normal distribution, the probability density function is given as
$p(x) = \exp(ax^2+bx+c) \quad (-\infty<x<\infty)$
where $a$, $b$, and $c$ are real. Suppose the mean value and standard deviation of $X$ are $x_0$ and $\sigma$, respectively. Find $a$, $b$, and $c$.

Answer:
The polynomial inside exponential can be modified as follows:
$$\begin{eqnarray} ax^2+bx+c &=& a\left[x^2+\frac{b}{a}x+\frac{c}{a}\right]  \\            &=& a\left[x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a}\right]  \\            &=& a\left[\left(x+\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a}\right]\\            &=& a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a} \end{eqnarray}$$
Thus, the probability density will become
$$\begin{equation} p(x) = \exp\left(-\frac{b^2-4ac}{4a}\right)\exp\left(a\left(x+\frac{b}{2a}\right)^2\right) \end{equation}$$
The probability density function with mean, $x_0$, and standard deviation, $\sigma$ forms
$$\begin{equation} p(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x-x_0)^2}{2\sigma^2}\right) \end{equation}$$
Compare this with the above expression.
$$x_0 = -\frac{b}{2a}, \quad -\frac{1}{2\sigma^2}=a$$
Therefore,
$$\begin{equation} a = -\frac{1}{2\sigma^2}, \quad b = \frac{x_0}{\sigma^2} \end{equation}$$
For the other part,
$$\frac{1}{\sqrt{2\pi}\sigma}= \exp\left(-\frac{b^2-4ac}{4a}\right)$$
Take the log.
$$\begin{eqnarray} -\log\sqrt{2\pi}\sigma&=& -\frac{b^2}{4a}+c  \\ c&=&-\frac{x^2_0}{2\sigma^2}-\log\sqrt{2\pi}\sigma \end{eqnarray}$$

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