Normal Distribution how $N(x-x_n|0,\sigma^2) = N(x |x_n,\sigma^2) $

Question

I read an expression

Could someone explain the step $N(t-t_n|0,\sigma^2) = N(t | t_n,\sigma^2) $ ?

Answer 1

It's just a translation along the axis.

The normal distribution of $t-t_n$ has a mean $0$ and variance of $\sigma^2$.

The normal distribution of $t$ will therefore have a mean of $t_n$ and variance of $\sigma^2$.

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$$\mathcal{N}(x\mid \mu, \sigma^2) = \mathcal{N}(x-a\mid \mu-a, \sigma^2)$$