Finding Marginal PDFs: Need Help With Integrating Complex Expressions?

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SUMMARY

The discussion centers on computing marginal probability density functions (PDFs) from a joint PDF, specifically integrating the joint PDF \( f_{X,Y}(x,y) \) with respect to \( y \) to find \( f_X(x) \). The integral involves complex expressions, but can be simplified using techniques such as completing the square and substitution. The standard integral of a normal distribution is utilized to facilitate the computation, leading to a clearer path for finding the marginal PDFs.

PREREQUISITES
  • Understanding of joint probability density functions (PDFs)
  • Knowledge of integration techniques, particularly with respect to multiple variables
  • Familiarity with normal distribution properties and integrals
  • Experience with mathematical substitutions in integrals
NEXT STEPS
  • Study the process of integrating joint PDFs to derive marginal PDFs
  • Learn about completing the square in the context of probability distributions
  • Explore substitution methods in integral calculus
  • Review the properties and applications of the normal distribution integral
USEFUL FOR

Mathematicians, statisticians, data scientists, and anyone involved in probability theory or statistical analysis will benefit from this discussion.

nacho-man
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Please see the attached image for my question. I don't understand how to compute the integral, is there some trick?
I do believe that
to find fx(x) I integrate the joint pdf, with respect to x with the bounds set as the range of Y. But this leaves me with a very complex integration
Similarly for fy(y). Is there some trick?

Any help is greatly appreciated.
 

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nacho said:
Please see the attached image for my question. I don't understand how to compute the integral, is there some trick?
I do believe that
to find fx(x) I integrate the joint pdf, with respect to x with the bounds set as the range of Y. But this leaves me with a very complex integration
Similarly for fy(y). Is there some trick?

Any help is greatly appreciated.
Hi nacho!

To find $f_X(x)$ you need to integrate $f_{X,Y}(x,y)$ with respect to y.

That is:
$$f_X(x) = \int_{-\infty}^{+\infty} f_{X,Y}(x,y) dy$$

The problem is of course that exponent, but you can complete the square, do a substitution, and use the standard integral of a normal distribution.

That is:
$$f_X(x) = \int_{-\infty}^{+\infty} f_{X,Y}(x,y) dy
= \int_{-\infty}^{+\infty} \frac{1}{\pi\sqrt 2} \exp\left(-(y + \frac 1 2 x \sqrt 2)^2 - \frac 1 2 x^2\right) dy$$

Can you substitute $w = y + \frac 1 2 x \sqrt 2$?

And use that $$\int_{-\infty}^{+\infty} \exp\left(-\frac 1 2 u^2\right) du = \sqrt{2\pi}$$?
 

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