Bivariate distribution question

[Longines]

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

 

[Jameson]

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.

 

[I like Serena]

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. ;)

 

[I like Serena]

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 $\mathbb E[XY]$?