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Let $X$ and $Y$ have a bivariate normal distribution with $\mu_X=5, \mu_Y=10, \sigma^2_X=1, \sigma^2_Y=25, \rho >0$.

If $P(4 < Y < 16|X=5)=0.9544$, I would like to find $\rho$.

I know that conditional marginals of a bivariate normal distribution are normal distributions. Given this knowledge, I can obtain the distribution $Y|X=5 \sim N(10,25(1-\rho^2)).$ However, integrating this pdf between $4$ and $16$ seems impossible. I have the following:

$f_{Y|X=5}(y)=(5\sqrt{2\pi(1-\rho^2)})^{-1}exp\{-(x-10)^2/(50(1-\rho^2))\}$, where $y \in R$.

$.9544=\int_4^{16}(5\sqrt{2\pi(1-\rho^2)})^{-1}exp\{-(x-10)^2/(50(1-\rho^2))\}dx,$

which does not seem possible to integrate. Is there a more efficient to solving this problem? Thank you.

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    $\begingroup$ How would you integrate a standard normal density between $a$ and $b$? $\endgroup$
    – Glen_b
    Nov 5, 2019 at 4:52
  • $\begingroup$ I see that the most common way is to use the Z table, but I was wondering if there was a more rigorous way to compute this. This problem is presented in my math stats class before we reach Z tables. $\endgroup$
    – Ron Snow
    Nov 5, 2019 at 4:53
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    $\begingroup$ You can't integrate it in "closed form" ... Doing this problem will rely on facts about the normal; in this case knowing how much of the probability is within 2 sd's of the mean. I bet you covered that much... $\endgroup$
    – Glen_b
    Nov 5, 2019 at 4:57
  • $\begingroup$ We did. Thank you for confirming my suspicions about the integration. I appreciate the guided thought! $\endgroup$
    – Ron Snow
    Nov 5, 2019 at 5:00
  • $\begingroup$ @Glen_b-ReinstateMonica why 2 sd's of the mean? It doesn't explicitly mentions that $\endgroup$ Feb 13, 2020 at 13:59

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Its integral cannot be expressed by elementary functions by definition. You'll use z-table. Let the RV represented by the marginal density of $Y$ given $X=5$ be denoted as $W$. We ask for $$\begin{align}P(4<W<16)&=P\left(\frac{4-10}{5\sqrt{1-\rho^2}}<Z<\overbrace{\frac{16-10}{{5\sqrt{1-\rho^2}}}}^a\right)\\&=P(-a<Z<a)=P(Z<a)-P(Z\leq-a)\\&=P(Z\leq a)-(1-P(Z\leq a))\\&=2P(Z\leq a)-1=0.9544\end{align}$$ Then, $P(Z\leq a)=0.9772\rightarrow a=2\rightarrow\rho=0.8$

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  • $\begingroup$ Thank you for your help. I assume this is the best way to compute the values, since an estimation is required any way. Thank you! $\endgroup$
    – Ron Snow
    Nov 5, 2019 at 4:54
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    $\begingroup$ @Edison yes it is the best way $\endgroup$
    – gunes
    Nov 5, 2019 at 5:17

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