MHB Bivariate distribution question

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The discussion focuses on solving a bivariate distribution problem involving integration from negative infinity to positive infinity. Participants discuss the method of completing the square to simplify the integral, leading to the marginal distribution function for X, f_X(x). One user mentions the potential use of polar coordinates but finds it unhelpful for this specific case. The conversation highlights the importance of recognizing algebraic techniques, such as completing the square, to facilitate solving the integral. Overall, the thread emphasizes collaborative problem-solving in understanding bivariate distributions.
Longines
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Hello all,

How would I do this question by hand?
I know I integrate from -infinity to +infinity for $f_x,y$, but I have no idea how to do it by hand! My algebra soup is bad, can someone please help me?

View attachment 3237P.S I heard some of my friends talking about some 'trick' you can do with the exponential part of the equation to solve it quicker.. but I don't know what they were talking about.

Thank you
 

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Hi there,

Usually with integrals in this form where you are integrating from $-\infty$ to $\infty$ I can find a way to do it by converting to polar coordinates, however I haven't found that yet. I think some users here might be able to see the proper substitution though - ZaidAlyafey, mathbalarka, MarkFL and ILS are all very skilled in integration. :)

I'll post back if I think of it and hope you get some more input soon.
 
Longines said:
Hello all,

How would I do this question by hand?
I know I integrate from -infinity to +infinity for $f_x,y$, but I have no idea how to do it by hand! My algebra soup is bad, can someone please help me?

View attachment 3237P.S I heard some of my friends talking about some 'trick' you can do with the exponential part of the equation to solve it quicker.. but I don't know what they were talking about.

Thank you

Hi Longines,

Let's start with $f_X(x)$.

$$f_X(x) = \int_{-\infty}^\infty f_{X,Y}(x,y)dy
=\int_{-\infty}^\infty \frac{1}{\pi\sqrt 2}\cdot e^{-x^2-\sqrt 2 xy-y^2}dy \tag 1
$$

Complete the square:
$$-x^2-\sqrt 2 xy-y^2 = -(y+\frac 12 \sqrt 2 x)^2-\frac 12 x^2 \tag 2
$$

Substitute (2) in (1):
$$f_X(x) = \int_{-\infty}^\infty \frac{1}{\pi\sqrt 2}\cdot e^{-(y+\frac 12 \sqrt 2 x)^2-\frac 12 x^2}dy
= \frac{e^{-\frac 12 x^2}}{\pi\sqrt 2}\int_{-\infty}^\infty e^{-(y+\frac 12 \sqrt 2 x)^2}dy
$$

Substitute $u=y+\frac 12 \sqrt 2 x$:
$$f_X(x)
= \frac{e^{-\frac 12 x^2}}{\pi\sqrt 2}\int_{-\infty}^\infty e^{-u^2}du
= \frac{e^{-\frac 12 x^2}}{\pi\sqrt 2} \cdot \sqrt \pi
= \frac{e^{-\frac 12 x^2}}{\sqrt{2\pi}}
$$

What do you think $f_Y(y)$ is? (Wondering)@Jameson: Sorry, no trick with polar coordinates. ;)
 
Hutchoo said:
Wow, thank you so much for this! This was exactly what I was looking for. I think the reason why I couldn't do it was because I didn't think of completing the square which is what's needed to solve the question.

Thank you!

Good!

Were you also able to find $f_Y(y)$ and $\mathbb E[XY]$?
 

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