3
$\begingroup$

I am trying to derive the expression for the PDF of a weighted mixture of n random variables. Let us taken $n=3$ and define $$X = \alpha_1 S_1 + \alpha_2 S_2 + \alpha_3 S_3$$ $$E[X^2] = 1$$ $s_1$, $s_2$, and $s_3$ are random variables which follow some distribution and have a joint distribution defined as $$f_{S_1S_2S_3}(s_1,s_2,s_3)$$ Then, $f_X(x;\alpha)$ is derived as: $$ S_3 = \frac{X - \alpha_1 S_1 - \alpha_2 S_2}{\alpha_3} $$ $$F_X(x;\alpha) = \frac{1}{\alpha_3} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\frac{x - \alpha_1 s_1 - \alpha_2 s_2}{\alpha_3}} f_{S_1 S_2 S_3}\left(s_1,s_2,s_3\right)ds_1 ds_2 ds_3$$ $$f_X(x;\alpha) = \frac{d}{dx}F_X(x;\alpha) = \frac{1}{\alpha_3} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{S_1 S_2 S_3}\left(s_1,s_2,\frac{x - \alpha_1s_1 -\alpha_2s_2}{\alpha_3}\right)ds_1 ds_2$$ Now, I would like to take the derivative of $f_X(x;\alpha)$ (I realize this may not have any statistical meaning, but I need to do it to understand the paper I am reading). The paper provides an answer for the derivative, but I am unsure how to derive it myself, which is where I need help. The answers are: $$ \frac{d}{d\alpha_1}f_X(x;\alpha)\bigg|_{\alpha_1=0, \alpha_2=0, \alpha_3=1} = -\int_{-\infty}^{\infty} \frac{df_{S_1S_3}(s_1,x)}{dx}s_1ds_1 $$ Similarly, $$ \frac{d}{d\alpha_2}f_X(x;\alpha)\bigg|_{\alpha_1=0, \alpha_2=0, \alpha_3=1} = -\int_{-\infty}^{\infty} \frac{df_{S_2S_3}(s_2,x)}{dx}s_2ds_2 $$

Can somebody provide how to calculate this derivative, or provide hints on which rules I need to apply to get to the results?

--- EDIT:Solution ---

1.) $s_1$, $s_2$, and $s_3$ are all continuous random variables over the domain of interest. Hence, we can move the derivative sign inside the integral. $$ \frac{d}{d\alpha_1}f_X(x;\alpha) = \frac{1}{\alpha_3} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{d}{d\alpha_1} \left[ f_{S_1 S_2 S_3}\left(s_1,s_2,\frac{x - \alpha_1s_1 -\alpha_2s_2}{\alpha_3}\right)\right]ds_1 ds_2 $$ 2.) Next, we apply the chain rule. $$ = \frac{1}{\alpha_3} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{-s_1}{\alpha_3} f'_{S_1,S_2,S_3}\left(s_1, s_2, \frac{x - \alpha_1s_1 -\alpha_2s_2}{\alpha_3} \right) ds_1 ds_2 $$ 3.) Next, we substitute the values of $\alpha_1=0$, $\alpha_2=0$, and $\alpha_3=1$ into the expression, yielding $$ = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} -s_1 f'_{S_1,S_2,S_3}\left(s_1, s_2, x \right) ds_1 ds_2 $$ 4.) Integrate over $\alpha_2$, yielding $$ = \int_{-\infty}^{\infty} -s_1 f'_{S_1,S_3}\left(s_1, x \right) ds_1 $$ $$ = -\int_{-\infty}^{\infty} \frac{df_{S_1,S_3}\left(s_1, x \right)}{dx} s_1 ds_1 $$

$\endgroup$
2
$\begingroup$

I'll give the rules you use/things to show to actually prove it. First, you should explain why you can pass the derivative under the integral and then apply the chain rule to the pdf. Then substitute in the given values for the alphas and evaluate the integral with respect to the s_i you didn't take the a_i derivative of. The result should fall out for both. I'd be happy to give more help.

$\endgroup$
  • 1
    $\begingroup$ thanks for the suggestion. I've worked out the steps as you mentioned and it as you said, the solution falls out (see above post edits). One additional question is, you'll notice between Steps 2 and 3, I switch from $X$ to $x$. Is this allowed, just merely a notational thing, or did I skip some step? $\endgroup$ – Kiran K. Jan 10 '15 at 16:53
  • $\begingroup$ I'm thinking it should be x the whole time (once we're looking at pdfs). In general, capital letters are used for the random variables themselves and lowercase letters for particular instances of the rvs. Since the pdf of the random variable gives the probability of the random variable to take a particular value, the pdf should only include the instance of the rv not the rv itself. $\endgroup$ – Nick Thieme Jan 10 '15 at 23:41
  • $\begingroup$ Yeah, I think you're correct. I thought about that as soon as I posted, but second opinion is always good. $\endgroup$ – Kiran K. Jan 11 '15 at 15:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.