Suppose that r.v $X$ and $Y$ are normally distributed and independent each other. Under which conditions $(X,Y)$ are bivariate normally distributed?
Now I want to calculate $\text{Prob}(X>a, Y<a, bX+cY<d)$, where $a,b,c,d$ are constant. Should I treat the third element as another r.v (let say $Z$) whose parameters are easily calculated and as a consequence, consider $Pr(X>a, Y<a, bX+cY<d)$ under a trivariate normal distribution perspective?
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1$\begingroup$ Possible duplicate of About Trivariate Normal Distribution $\endgroup$– Xi'anCommented Oct 23, 2015 at 17:24
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1$\begingroup$ no,the question is different (it seems) $\endgroup$– AndreaCommented Oct 23, 2015 at 17:27
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Well if $X$ and $Y$ are independent and each one is normally distributed, then $(X,Y)$ is also bivariate normal (with $Cov(X,Y)=0$). And no, I do not suggest using a trivariate normal distribution because the distribution of $(X,Y, bX+cY)$ is degenerate, in the sense that its support lies in a 2 dimensional subspace of $\mathbb R^3$.
You just need to integrate over the pdf of the bivariate normal (which due to Fubini's theorem and independence is really simple). For example if $b>0$ then $bX+cY < d$ is equivalent to $X < \frac{d-cY}{b}$. Then you just integrate:
$$\Pr[X>a, Y<a, bX+cY<d] = \int_{-\infty}^a\int_{a}^{\frac{d-cy}{b}} \mathbf{1}_{\left\{\frac{d-cy}{b}\geq a\right\}} f_X(x)f_Y(y) dx dy$$
Here $f_X$ (resp. $f_Y$) is the density of the distribution of $X$ ($Y$) and $\mathbf{1}_{\{\}}$ is the indicator function.
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$\begingroup$ thank you! what about if b or c are negative? I take advantage of your expertise for another question: as you don't suggest to use a trivariate normal distribution,which package of R do you suggest in this case (I've thought of mvtnorm in case of trivariate distribution).Many thanks $\endgroup$– AndreaCommented Oct 23, 2015 at 17:37
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$\begingroup$ Part1) Well if b is negative then you just flip the inequality around and it holds that $X < \frac{d-cY}{b}$. I will let you figure out on your own what this means regarding the integration region! Also to be honest I would probably derive an analytic expression for this case (at least get rid of the inner integral) and then just write my own R function. But I am sure there are also numeric integration packages you can use in R (but not sure which or how you would specify the information). $\endgroup$– airCommented Oct 23, 2015 at 17:45
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$\begingroup$ Part 2) But for such a low dimension you can actually do it via Monte Carlo if the integrals scare you! So you just generate independent random variables $X_i$, $Y_i$ according to the distribution of $X$ and $Y$. And you do that many times (e.g. $m$ times) so that you get pairs $(X_i,Y_i)$ for $i=1,\dotsc,m$. Now for each of these pairs you check whether all 3 conditions ($X_i >a$, $Y_i < a$ and $bX+cY<d$) are fullfilled. Now the probability you are interested in is just the proportion of pairs which fullfill this property. $\endgroup$– airCommented Oct 23, 2015 at 17:47