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A problem is this: The probability density function of the univariate Gaussian with mean $ μ $ and variance $σ2, N(μ,σ2)$:

$$f_x(x) = \frac{1}{\sqrt(2*pi*σ2)} * e^-(x-μ)^2/(2*σ2)$$

The pdf of a Gaussian random variable X is given by: $$ f_x(x) = \frac{n}{(3*\sqrt(2*pi))} * exp(-(n^2(x-2)^2)/18)$$.

What is the mean and variance of X?

I got the mean right: 2, but the variance wrong:9.

Why is 9 not the right answer here? I thought it would be $3^2$?

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You omit $n$ in the expression. Your PDF can be written as $$f_X(x)=\frac{1}{\sqrt{2\pi}(3/n)}\exp(-(x-2)^2/(2(3/n)^2))$$ where $\sigma_2=3/n$.

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  • $\begingroup$ I see, thanks Gunes! $\endgroup$ – speedy_catch Feb 11 at 13:16

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