# Marginal PDF

#### nacho

##### Active member
Please refer to the attached image.

Is there a way to simplify this question? It looks really messy but I have a feeling there is some nifty way around it. Surely they don't want us to integrate that entire function.

I also have no idea about part d)
how can some values of lambda result in X and Y being independent, and others not?

Any help appreciated.

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#### chisigma

##### Well-known member
Please refer to the attached image.

Is there a way to simplify this question? It looks really messy but I have a feeling there is some nifty way around it. Surely they don't want us to integrate that entire function.

I also have no idea about part d)
how can some values of lambda result in X and Y being independent, and others not?

Any help appreciated.
Before to try some 'brute force approach' my be it is useful to observe that the given joint distribution...

$\displaystyle f(x,y) = \frac{1}{2\ \pi\ \lambda^{3}}\ e^{- \frac{1}{2\ \lambda^{4}}\ \{(x-\lambda)^{2} + (y-\lambda)^{2} - 2\ \sqrt{1-\lambda^{2}}\ (x-\lambda)\ (y-\lambda)\}}\ (1)$

... is a particular case of the bivariate normal distribution...

$\displaystyle f(x,y) = \frac{1}{2\ \pi\ \sigma_{x}\ \sigma_{y}\ \sqrt{1- \rho^{2}}}\ e^{- \frac{z}{2\ (1-\rho^{2})}}\ (2)$

... where...

$\displaystyle z = \frac{(x-\mu_{x})^{2}}{\sigma_{x}^{2}} + \frac{(y-\mu_{y})^{2}}{\sigma_{y}^{2}} - 2\ \frac{\rho\ (x - \mu_{x})\ (y-\mu_{y})}{\sigma_{x}\ \sigma_{y}}\ (3)$

... and $\displaystyle \mu_{x}= \mu_{y} = \sigma_{x} = \sigma_{y} = \sqrt{1-\rho^{2}} = \lambda$...

Kind regards

$\chi$ $\sigma$

#### nacho

##### Active member
thanks chi

I am unsure what to do with this information, how am I supposed to utilise it?

I have a feeling that you want me to use known properties of a bivariate distribution, and that the function will have the same properties, just with some transformations?

could also you suggest what I area I study in order to solve questions like these?

#### chisigma

##### Well-known member
Before to try some 'brute force approach' my be it is useful to observe that the given joint distribution...

$\displaystyle f(x,y) = \frac{1}{2\ \pi\ \lambda^{3}}\ e^{- \frac{1}{2\ \lambda^{4}}\ \{(x-\lambda)^{2} + (y-\lambda)^{2} - 2\ \sqrt{1-\lambda^{2}}\ (x-\lambda)\ (y-\lambda)\}}\ (1)$

... is a particular case of the bivariate normal distribution...

$\displaystyle f(x,y) = \frac{1}{2\ \pi\ \sigma_{x}\ \sigma_{y}\ \sqrt{1- \rho^{2}}}\ e^{- \frac{z}{2\ (1-\rho^{2})}}\ (2)$

... where...

$\displaystyle z = \frac{(x-\mu_{x})^{2}}{\sigma_{x}^{2}} + \frac{(y-\mu_{y})^{2}}{\sigma_{y}^{2}} - 2\ \frac{\rho\ (x - \mu_{x})\ (y-\mu_{y})}{\sigma_{x}\ \sigma_{y}}\ (3)$

... and $\displaystyle \mu_{x}= \mu_{y} = \sigma_{x} = \sigma_{y} = \sqrt{1-\rho^{2}} = \lambda$...
The main reason why I remember the normal bivariate distribution is that we can use its very comfortable properties...

Bivariate Normal Distribution -- from Wolfram MathWorld

Regarding the points a) we have that the marginal distribution function of the X is...

$\displaystyle f_{x} (x) = \int_{- \infty}^{+ \infty} f(x,y)\ dy = \frac{1}{\sigma_{x}\ \sqrt{2\ \pi}}\ e^{- \frac{(x - \mu_{x})^{2}}{2\ \sigma_{x}^{2}}} = \frac{1}{\lambda\ \sqrt{2\ \pi}}\ e^{- \frac{(x - \lambda )^{2}}{2\ \lambda^{2}}}\ (1)$

Now the point b) is direct consequence of (1)... why?...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
The main reason why I remember the normal bivariate distribution is that we can use its very comfortable properties...

Bivariate Normal Distribution -- from Wolfram MathWorld

Regarding the points a) we have that the marginal distribution function of the X is...

$\displaystyle f_{x} (x) = \int_{- \infty}^{+ \infty} f(x,y)\ dy = \frac{1}{\sigma_{x}\ \sqrt{2\ \pi}}\ e^{- \frac{(x - \mu_{x})^{2}}{2\ \sigma_{x}^{2}}} = \frac{1}{\lambda\ \sqrt{2\ \pi}}\ e^{- \frac{(x - \lambda )^{2}}{2\ \lambda^{2}}}\ (1)$

Now the point b) is direct consequence of (1)... why?...
The knowledge of $\displaystyle f_{x} (x)$ and the following useful article...

http://mpdc.mae.cornell.edu/Courses/MAE714/biv-normal.pdf

... permits us to find the conditional distribution...

$\displaystyle f_{Y|X=x} (y) = \frac{f(x,y)}{f_{x}(x)} = \frac{1}{\sqrt{2\ \pi}\ \sigma_{y}\ \sqrt{1 - \rho^{2}}}\ e^{- \frac{1}{2\ \sigma_{y}^{2}\ (1-\rho^{2})}\ \{y - \mu_{y} - \rho\ \frac{\sigma{y}}{\sigma_{x}}\ (x-\mu_{x})\}^{2}} = \frac{1}{\sqrt{2\ \pi}\ \lambda^{3}}\ e^{- \frac{1}{2\ \lambda^{4}}\ \{y - 2\ \lambda - \lambda\ (x-\lambda)\}^{2}}\ (1)$

Now the (1) is a standard normal ditribution with mean $\displaystyle \mu_{y} + \rho\ \frac{\sigma_{y}}{\sigma_{x}}\ (x - \mu_{x})$ and variance $\displaystyle (1- \rho^{2})\ \sigma_{y}^{2}$ so that is...

$\displaystyle E \{Y|X=x\} = \lambda + \sqrt{1 - \lambda^{2}}\ (x - \lambda)\ (2)$

... and the point b) is answered... the remaining point c) and d) shouldn't be too difficult to attack at this point... expecially the point d)!...

Kind regards

$\chi$ $\sigma$

#### nacho

##### Active member
point of clarification - how would you describe the parameters of our joint PDF in terms of the bivariate normal distribution?

also, is it independent for $\lambda$ = 1? for part d)

Paramaters of distributions always confuse me, I lost a huge bulk of marks on my mid-term because I don't understand what is trying to be said/communicated.

Would you be able to explain that?
Thanks!

edit: also for part a) when finding the marginal distributions, is it sufficient for us to say $f_{X}(x) = \int...dy = ...$ or do we actually have to show the integration? ie, is this just a property we can use, or one which we must derive?

for part b)
i got a different answer from you, and i dont know what I did wrong

so $f_{X}(x)$ = $\frac{1}{\lambda \sqrt{2\pi}}$ $e^{(\frac{-(x-\lambda)^2)}{2(\lambda)^2}}$

and $f_{X,Y}(x,y)$ = $\frac{1}{2 \pi (\lambda)^3}$ $e^{...}$

and
$\frac{f_{X,Y}(x,y)}{f_X(x)}$ should at least have $(\lambda)^2$in the denominator, as opposed to a $\lambda^4$ ?
additionally, my exponential was a different power from yours, i don't know how you simplified yours

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