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How can I convert a the pdf of a normal distribution that it N(t,1), but integrated from 0 to infinity, to the standard normal. I found that the former is equal to 1- ϕ(-t) but i cant figure how this works. By convert i want to represent N(t,1) in terms of the CDF of N(0,1)

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Let $X$ be normal mean $\mu$, standard deviation $\sigma$. Then $$F_X(x)=\Pr(X\le x)=\Pr\left(Z\le\frac{x-\mu}{\sigma}\right),\tag{1}$$ where $Z$ is standard normal.

Yours is the case $\mu=t$, $\sigma=1$. But we might as well continue with the general case. To find the pdf of $X$, from (1) we obtain that $$F_X(x)=\int_{-\infty}^{(x-\mu)/\sigma}\frac{1}{\sqrt{2\pi}}e^{-z^2/2}\,dz.$$ Differentiate with respect to $x$, using the Fundamental Theorem of Calculus. We get $$f_X(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/(2\sigma^2)}.$$ More briefly, $$f_X(x)=\frac{1}{\sigma}f_Z((x-\mu)/\sigma).$$

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