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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. $$\frac{1}{\sqrt[]{2\pi\sigma^2t}\ }\exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right)$$

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

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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 $t > 0$.

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    $\begingroup$ The variance is $\sigma^2 t$ $\endgroup$
    – Gregory
    Commented Feb 11, 2021 at 17:00
  • $\begingroup$ Could you explain how to achieve that? When using the original proof of normal pdf, I face some conflicts? Could you give some hints about the proof? $\endgroup$ Commented Feb 11, 2021 at 17:01

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