# Mean and Variance of Normal Distribution with parameters

I see a special pdf function on textbook. I guess that it is the pdf of normal distribution but with a strange parameter t. Is my guessing right? And could someone explain why this pdf has a t? I hope to calculate the mean and variance to see whether it is normal distribution. I can remember that $$\int_{-\infty}^{\infty}\frac{1}{\sqrt[]{2\pi}}exp(-\frac{x^2}{2})dx=1$$. But I don't know how to work in the case with this t. Could someone give some hints about it? Thank you very much. $$\frac{1}{\sqrt[]{2\pi\sigma^2t}\ }\exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right)$$

• Is there a typo? Should $t$ be under the square root? Commented Feb 11, 2021 at 16:44
• Sorry about that. I think I misread the formula. It is a typo. Commented Feb 11, 2021 at 16:59

This is just a case of the usual normal distribution in which the mean and variance parameters are linked through a third parameter $$t$$. The usual parametrization for a normal density is $$f_X(x) = \frac{1}{\sqrt{2\pi \sigma^2}} \exp \left(-\frac{(x-\mu)^2}{2\sigma^2}\right).$$ Here the mean is $$\mu$$ and the variance is $$\sigma^2$$. Now in your case, the mean is $$\mu t$$ and the variance is $$\sigma^2 t$$. Note we require $$t > 0$$.
• The variance is $\sigma^2 t$ Commented Feb 11, 2021 at 17:00