Let $X_1, X_2, \cdots, X_m$ be random samples from a normal distribution $N(\theta_1, \theta_2)$.


$$ L(\theta_1,\theta_2) = P(X_1=x_1;X_2=x_2;\cdots;X_n=x_m) = \prod_{i=1}^{m} \dfrac{1}{\sqrt{2\pi\theta_2}}{\text{exp}}{\Big[ -\dfrac{ (x_i-\theta_1)^2 }{2\theta_2} \Big]} \tag{1} $$

To relate to my doubt, I will shrink sample size to $m=2$. That is, we have $X_1,X_2$. Then,

$$ L(\theta_1,\theta_2) = f_{X}(x_1)f_{X}(x_2) = P(X_1=x_1;X_2=x_2) = \prod_{i=1}^{2} \dfrac{1}{\sqrt{2\pi\theta_2}}{\text{exp}}{\Big[ -\dfrac{ (x_i-\theta_1)^2 }{2\theta_2} \Big]} \tag{2} $$

But as per product distributions wiki, if $X_1$ and $X_2$ are two independent continuous random variables, both described by probability density function $f_{X}$, then the probability density function of $Z = X_1X_2$ would be

$$ f_Z(z) = \int\limits_{-\infty}^{\infty}f_{X}(x_1)f_{X}(z/x_1)\dfrac{1}{|x_1|}dx_1 \tag{3} $$

My questions:
1) Does eq.{2} also represent case $Z = X_1X_2$? If so, how do I reduce {3} to {2}?
2) If not, how are {2} and {3} are not related?
3) Is product distribution of $Z=X_1X_2$ same as joint probability distribution $f_{X}(x_1,x_2)$? What are the relation between the two.

A related question was posted here, from which this doubt arose. Now I am confused, why eq {3} did not interfere in my MLE case.

I took above MLE example from here, page 260.


1) Does eq.{2} also represent case $Z=X_1X_2$?

Absolutely not. There's no reason for you to think that. The product of densities (or product of CDFs) is not the density of product of random variables.

I don't see anywhere in the previous post you linked that suggested that.

  • $\begingroup$ I am somehow naively getting confused between samples $X_1, X_2$, and tend to treat them as individual independent RVs because, we multiply their density functions to find a collective joint density function (so implicitly going backwards, since I associated a density function to each $X_m$, I tend to treat it as a RV). $\endgroup$ – Parthiban Rajendran Oct 18 '18 at 16:01

The product distribution is the distribution of the random variable $Z = X \times Y$.

The joint distribution of $X,Y$ is the distribution of the random vector $(X,Y)$.

The two are not the same at all.

  • $\begingroup$ So in my MLE example above, we take a joint distribution? $\endgroup$ – Parthiban Rajendran Oct 18 '18 at 16:00
  • $\begingroup$ Yes. The joint distribution represents the "likelihood" (in a sense) of getting that sample, hence why its used for MLE $\endgroup$ – Xiaomi Oct 18 '18 at 16:01
  • $\begingroup$ Suppose X and Y are 2 independent identically distributed RVs. Then if we say, the joint pdf is $f_{XY}(x,y) = f_{X}(x)f_{Y}(y)$, it would mean backwards, X and Y were RVs, so we used individual density functions to get joint.Similarly here, we use $f_{X}(x_1)f_{X}(x_2)$, so why we would not say, we treated $X_1, X_2$ as RVs? $\endgroup$ – Parthiban Rajendran Oct 18 '18 at 16:07
  • $\begingroup$ Using either is fine. You need to use the joint, but if they are independent it is simpler to use the marginals and multiply them together. $\endgroup$ – Xiaomi Oct 18 '18 at 16:09
  • $\begingroup$ When I say the sample set is drawn from same distribution, how do the samples become independent? Does my example implicitly imply sampling with replacement? If so, without replacement, MLE would be different? I am sorry a base question is raising many associated doubts. $\endgroup$ – Parthiban Rajendran Oct 18 '18 at 16:15

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