I am reading through a derivation of the bivariate normal distribution, (from the US Defence Department!) and came across an expression that I can't understand.

The derivation starts off with the observation that the total area, $A$, under the curve of the distribution is $1$, since it is a probability distribution. Also, the distribution can be expressed as a differential equation:

$$dy=-k \ xy \ dx \tag{1}$$

$$y= Ce^{-k{x^2\over2}} \tag{2}$$

$$A= C\int_{-\infty}^{\infty}e^{-k{x^2\over2}} \tag{3}$$

Using integration by parts, the text proceeds to evaluate

$$A= C|e^{-k{x^2\over2}}|_{-\infty}^{\infty}+C\int_{-\infty}^{\infty}x^2e^{-k{x^2\over2}} \tag{4}$$

(I skipped some steps for brevity. Please tell me if that creates confusion.)

Then the text says the following line:

Since the first expression takes an indeterminate form, new numerators and denominators are obtained by independent derivatives, and the limiting value of the expression then becomes zero. Since, by definition, the n-th moment of a frequency distribution is defined as

$$\text {n-th moment} = \int_{b}^{a}x^n f(x)dx$$

and since from fundamental principles the standard deviation is the square root of the second moment about the mean, then it follows, considering $(3)$, that the second expression in $(4)$ becomes $$A=Ak\sigma_X^2 \tag{5}$$

I think I understand how we got rid of the first expression $C|e^{-k{x^2\over2}}|_{-\infty}^{\infty}$ in $(4)$, but what I don't understand is, how does this last step, $(5)$, follow from the previous steps? Also, where did the $C$ go?


$C$ is part of $f(x)$ as can be seen in step $(3)$ if one would just put it in front of the integral sign, and is therefore part of the variance, $σ_X^2$. $A$ is just $1$. The point of this step is to prove that $k={1\over σ_X^2}$


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