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 $\mathbb E[XY]$?

 

[blue_raver22]

for reaching out with your question. Bivariate distribution questions can be solved by hand using techniques from multivariate calculus and algebra. The first step is to determine the joint probability density function (PDF) of the two variables, denoted as $f_{x,y}$. This can be done by integrating the joint PDF over the entire range of both variables, in this case from -infinity to +infinity. This will give you the marginal distribution of each variable.

Next, you can use the joint PDF to calculate the correlation coefficient, which measures the strength and direction of the relationship between the two variables. This can be done by finding the covariance of the two variables and dividing it by the product of their standard deviations.

As for the "trick" your friends mentioned, it is possible they were referring to using the properties of the exponential function to simplify the calculations. However, it is important to have a solid understanding of the underlying concepts and equations before attempting any shortcuts.

I suggest reviewing the fundamentals of bivariate distributions and seeking help from a mathematics tutor or online resources if needed. With practice and patience, you will be able to solve bivariate distribution questions by hand. Good luck!