Integrating Normal Distribution with $1$, $x$, and $x^2$

Question

I am taking an introductory Machine Learning class but it has been a while since I took calculus. I am having a difficulty in understanding the following equations.

So we are currently talking about the Gaussian Distribution: $$\int_{-\infty}^{\infty} N(x|\mu,\sigma^2)dx = 1$$

This makes sense to me as I understand that the sum of the probability distribution adds up to one.

Where I start to struggle is the following: $$\int_{-\infty}^{\infty} N(x|\mu,\sigma^2)xdx = \mu$$ I dont understand how adding all the x values from $-\infty$ to $\infty$ gives us the mean.

And this is the next step in the equations, which once again, I cant follow but I believe If I can get the earlier one, this one will be easy as well. $$\int_{-\infty}^{\infty} N(x|\mu,\sigma^2)x^2dx = \mu^2 + \sigma^2$$

Sorry for the trivial looking question. If I can wrap around the intuitive meanings the rest should be easier I feel. The content is from Pattern Recognition and Machine Learning.

Thank you

Answer 1

In you second equation it is important to notice that you are weighting $x$ by the probability of getting $x$, so the most likely values are going to contribute more to the sum, than the less likely values. That's why

$$ \int_{-\infty}^{+\infty}{\rm d}x~xN(x|\mu,\sigma^2) = \mu $$

All this is saying is that average value of $x$ is $\mu$.

As for the last equation, you're ultimately calculating the integral

$$ \int_{-\infty}^{+\infty}{\rm d}x~(x-\mu)^2N(x|\mu,\sigma^2) $$

which just represents the average (squared) distance from the all points to the mean value $\mu$

Answer 2

just write it out: $$ \int_{-\infty}^{\infty} N(x|\mu,\sigma^2)dx = \int_{-\infty}^{\infty} \dfrac {1}{\sqrt {2\pi} \sigma} e^{- \frac {(x-\mu)^2}{2\sigma^2}}dx=1 $$ and $$ \int_{-\infty}^{\infty} N(x|\mu,\sigma^2)xdx =\int_{-\infty}^{\infty} \dfrac {1}{\sqrt {2\pi} \sigma} e^{- \frac {(x-\mu)^2}{2\sigma^2}}xdx $$ change variables $z=\dfrac {x-\mu}{\sigma}$, we get $$ =\int_{-\infty}^{\infty} \dfrac {1}{\sqrt {2\pi}} e^{- \frac {z^2}{2}}(\sigma z + \mu )dz=\sigma\int_{-\infty}^{\infty} \dfrac {1}{\sqrt {2\pi}} e^{- \frac {z^2}{2}} zdz+\mu\int_{-\infty}^{\infty} \dfrac {1}{\sqrt {2\pi}} e^{- \frac {z^2}{2}} dz $$ the first integral in the right hand side is zero since the integrand is odd; the second is $\mu\int_{-\infty}^{\infty} N(x|0,1)dx=\mu$. Then we are done.

You can do the next case on your own.