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I've been dealing with this proability problem, any advice would be really appreciated:

Let $(X,Y)$ be a normal bivariate random vector, such that: $\mathbb{E}[X]=2$, $Var[X]=4$, and $\mathbb{E}[Y|X=x]=2+x$ and $Var[Y|X=x]=5$. Calculate a) $\mathbb{P}(Y>5)$ and b) $\mathbb{E}[Y^2|X=7]$

My attempt:

I have no clue on the a) part, since I can't know how Y distributes, because I don't know if X,Y are independent.

For part b) I used the following, relative to conditional variance:

$Var[Y|X=x]=\mathbb{E}[Y^2|X=x]-(\mathbb{E}[Y|X=x])^2$

So I only computed when $X=7$ and obtained:

$Var[Y|X=7]=\mathbb{E}[Y^2|X=7]-(\mathbb{E}[Y|X=7])^2$

$5=\mathbb{E}[Y^2|X=7]-(2+7)^2=\mathbb{E}[Y^2|X=7]-81$

So:

$\mathbb{E}[Y^2|X=7]=86$. But I'm not sure if I did something wrong. Thanks in advance.

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  • $\begingroup$ $X,Y$ clearly are not independent. Also please check the expression of $E[Y \mid X=x]$ again, it is probably wrong. $\endgroup$ – air Dec 11 '16 at 5:52
  • $\begingroup$ I think I've thought that, it makes difficult to calculate $P(Y>5)$ right? $\endgroup$ – User117E29 Dec 11 '16 at 5:54
  • $\begingroup$ Damn, I have this exercise as a past final exam on a course of probability, and It says exactly that, i.e. the part of E[Y|X=x]=2+x. Maybe it could be a mistake? $\endgroup$ – User117E29 Dec 11 '16 at 5:56
  • $\begingroup$ Oh that makes a lot of sense, in your post above you also have a random $y$ floating around so that I thought it's $E[Y|X=x] = 2 + xy$. $\endgroup$ – air Dec 11 '16 at 7:24
  • $\begingroup$ Ohhh I seee. Jaja. It was my mistake, Im mexican so Im translating it to English, That "y" must be "and" (the conjunction in spanish) sorry about That $\endgroup$ – User117E29 Dec 11 '16 at 7:29
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Step 1: Conditionally on $X$

First work conditionally on $X$. Here you use the property of a normal distribution, that if $(X,Y)$ are jointly normal, then $Y \mid X=x$ is also normal. Hence, since you know the conditional mean and variance you know that:

$$ Y \mid X=x \sim \mathcal{N}(2+x, 5)$$

Based on the above it is easy to calculate $ \mathbb P[ Y > 5 \mid X=x]$.

Step 2: Marginalize over $X$

Now notice that $\mathbb P[Y > 5 \mid X]$ is a function of $X$. Again by properties of multivariate normal you know that $X$ is normal and since you know its mean and variance, you know the distribution of $X$. Now just integrate over $X$, i.e.

$$ \mathbb P[Y > 5] = \mathbb E[ \mathbb P[Y >5 \mid X]] $$

I'll leave the calculations to you.

Alternative way:

Again for a normal, there is a very explicit expression for the conditional distribution, see e.g. the wiki article. From this and your information, you can quickly figure out what the variance and mean of $Y$ are, and since you know $Y$ must be normal too, you know its distribution.

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  • $\begingroup$ Thanks a lot, air. Two doubts. When I marginalize over X (after doing step 1), I have to integrate $1-\Phi((3-x)/5)$ over -infinity to infinity? Isn't that an indeterminate form? And fot the Alternative form, you mean the formula that states that $E[Y|X=x]=\mu_y + \rho\sigma_y /\sigma_x (X-\mu_x)$? $\endgroup$ – User117E29 Dec 11 '16 at 23:43
  • $\begingroup$ Yes for your 2nd question (and there's a similar equation for $Var(Y | X=x)$ which you also need. For the first one, the integral should still exist (remember to include the density of $X$ in the integration). I did not try calculating by the 1st way, but maybe by Fubini or so it works out. Also check results with both methods. $\endgroup$ – air Dec 12 '16 at 1:13
  • $\begingroup$ Thanks a lot, Air. I really appreciate your help and patience. $\endgroup$ – User117E29 Dec 12 '16 at 7:50
  • $\begingroup$ P.d. sorry for my poor English $\endgroup$ – User117E29 Dec 12 '16 at 7:51

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