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Consider a normally distributed random variable $X\sim N(\mu,\sigma^2)$.

Since $\Pr[X\leq x]=\Pr[(X-\mu)/\sigma\leq (x-\mu)/\sigma]$, the following equality holds: $$\Phi(x|\mu, \sigma^2)=\Phi((x-\mu)/\sigma\;|\;0,1)$$ where $\Phi(\cdot|\mu,\sigma^2)$ is a CDF of a normal distribution with mean $\mu$ and variance $\sigma^2$.

That is, we can express a normal CDF using the standard normal CDF.

Here, I thought that an equality $\Pr[X= x]=\Pr[(X-\mu)/\sigma= (x-\mu)/\sigma]$ may also hold and thus $\phi(x|\mu,\sigma^2)=\phi((x-\mu)/\sigma|0,1)$.

However, I checked that this is not true. (In R, dnorm(3, 2, 3) $\neq$ dnorm((3-2)/3)).

So, why does the equality not hold in the PDF case? and is it possible to express a normal PDF using the standard normal PDF?

MinChul Park
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    To move from the CDF to the PDF, you need to take derivatives with respect to $x$. This means you need to apply chain rule on the RHS. Then we should expect that dnorm(3,2,3)=dnorm(1/3)/3 – user1848065 May 09 '23 at 02:46
  • @user1848065 great answer! thank you! – MinChul Park May 09 '23 at 03:53
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    This means you forgot the Jacobian, which is $1/\sigma$ in that case. – Xi'an May 09 '23 at 04:21
  • @Xi'an That is correct. Thank you. – MinChul Park May 09 '23 at 04:23
  • https://stats.stackexchange.com/search?q=pdf+transform Often, plotting things resolves such questions. Looking at curve(dnorm((x-2)/3), -5, 10); curve(dnorm(x,2,3), add = TRUE, col = "Red") will make it clear what's going on and takes even less effort than asking the question! – whuber May 09 '23 at 18:32

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