Hello all, I was trying to find the probability of a Joint distribution function given by the following double integral but was unable to get the right value $$ P = \int_{-\infty}^{\infty} \int_{-\infty}^{y+2.6} f_{X,Y}(x,y) dx dy = 0.9467$$ where $$ f_{X,Y}(x,y) = \frac{1}{A} \exp^{-\frac{1}{2}(x^2+y^2+2rxy)} $$ for $ r = 0.23 $ ; $ \{X,Y\}\in Real $ and $X - Y \leq 2.6$
I was trying the following script
r = 0.23;
fxy = (1/A)*Exp[(-1/2)*(x^2 + y^2 + 2*r*x*y)];
Inum = Integrate[fxy, {x, -Infinity, y + 2.6}, {y, -Infinity, Infinity}];
Aval = NSolve[Inum == 1, A];
fxy = fxy /. Aval;
P = Integrate[fxy, {x, -Infinity, y + 2.6}, {y, -Infinity, Infinity}]
for which I got the output shown below ![enter image description here](https://community.wolfram.com//c/portal/getImageAttachment?filename=output.PNG&userId=1809845)
I will appreciate if anyone can help me figure out the issue here to get the exact value of P as given above. Thanks in advance.