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I have the following problem. I am reading the paper which uses this identity for a proof, but I can't see why or how to prove its true. Can you help me?

\begin{align} \int_{x_{0}}^{\infty} e^{tx} n(x;\mu,\nu^2)dx &= e^{\mu t+\nu^2 t^2 /2} N(\frac{\mu - x_0 }{\nu} +\nu t ) \end{align}

where $n(\cdot)$ is the normal pdf with mean $\mu$ and variance $\nu^2$. $N(\cdot)$ refers to the normal cdf.

\begin{align} \int_{x_{0}}^{\infty} e^{tx} n(x;\mu,\nu^2)dx &= \int_{x_{0}}^{\infty} e^{tx} \frac{1}{\nu \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2 \nu^2} } dx \\ &\int_{x_{0}}^{\infty} \frac{1}{\nu \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2 \nu^2} + tx } dx \\ &\int_{x_{0}}^{\infty} \frac{1}{\nu \sqrt{2\pi}} e^{-\frac{x^2 -2x\mu +\mu^2 +2\nu^2tx}{2 \nu^2}} dx \\ \end{align}

and I'm stuck here Thank you.

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1 Answer 1

up vote 1 down vote accepted
  1. Substitute the expression for the Normal pdf.
  2. Gather together the powers of $e$.
  3. Complete squares in the exponent of $e$ to get the square of something plus a constant.
  4. Take the constant powers of $e$ out of the integral.
  5. Change variables to turn the integral into a integral of the standard normal pdf from $-\infty$ to some number $a$.
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