0

I have problems when I get the integral $$\int_{0}^{\infty}\frac{e^{-(\frac{x^{2}}{2u}+u)}}{\sqrt{u2\pi}} du$$ what should I do? Or this result is wrong?

BlueRedem1
  • 506
  • 1
    Your integral diverges: the integrand gets large as $u \to \infty$. Perhaps that should be $-u$ in the exponent? – Robert Israel Apr 27 '20 at 23:59
  • @RobertIsrael Sorry I forgot to write the parenthesis. But I don't know if my result is right or wrong because I can't solve that integral... – BlueRedem1 Apr 28 '20 at 00:05
  • 1
    If the conditional distribution of $X$ is standard normal when given $\Sigma^2=u$, then $X$ is independent from $\Sigma^2$. Did you instead mean $X\mid\Sigma^2{=}u\sim\mathcal {N}(0,u)$ ? – Graham Kemp Apr 28 '20 at 00:13
  • @GrahamKemp the exercise says that is a Standard normal. I tried to complete another normal distribution to get rid of the integral but I can't do that. If the conditional distribution is a standard normal then both variables are independent? I haven't seen that result yet – BlueRedem1 Apr 28 '20 at 00:18
  • 1
    If $X\mid Y=y$ has a distribution whose parameters do not depend on $y$, then $X$ and $Y$ are independent. If the exercise does indeed say $X\mid\Sigma^2{=}u\sim\mathcal N(0,1)$, then: $$f_{X\mid\Sigma^2}(x\mid u)=\dfrac{\mathrm e^{-x^2/2}}{\surd(2\pi)}$$ – Graham Kemp Apr 28 '20 at 00:21
  • @GrahamKemp Sorry, I wrote that wrong. You were right. But I worked with $N\sim(0,u)$ and got that result. I'm stuck on that point – BlueRedem1 Apr 28 '20 at 00:27
  • Well, then it looks correct, but I cannot see that integrating in terms of standard functions... – Graham Kemp Apr 28 '20 at 00:41
  • Your integral is correct. See here for how to evaluate it (or to avoid evaluating it and find the characteristic function instead). You will get the Laplace distribution. – Maxim Apr 28 '20 at 05:48

0 Answers0