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I am trying to find the CDF of $Z=aX+bY$ whereby $X$ and $Y$ are random variables and $a$ and $b$ are positive integers. Given that the CDF of $Z$ is:

$$F_Z(z)=\int F_X\left(\frac{z-by}{a}\right)f_Y(y)dy$$

Given that $X$ is uniform distribution over $[0,1]$ with: $$f_X \left(x \right)=\frac{1}{\sigma }\left [ H\left ( x \right ) -H\left ( x-\sigma \right )\right ]$$ $$F_X \left(x \right)=\frac{x}{\sigma }\left [ H\left ( x \right ) -H\left ( x-\sigma \right )\right ]$$

And $Y$ is Weibull distribution over $[0, \infty)$ with:

$$f_Y \left(x \right)=\frac{\beta }{\sigma }\left (\frac{x}{\sigma } \right )^{\beta -1} e^{-\left ( \frac{x}{\sigma } \right )^{\beta }}$$ $$F_Y \left(x \right)=1- e^{-\left ( \frac{x}{\sigma } \right )^{\beta }}$$

where $\beta, \sigma >0$ in all the expressions above.

I tried deriving $F_Z(z)$ following some videos online but no luck when I check against numerical CDF generated using Monte Carlo (random numbers). Can help me determine the limits and explain just once on how to solve the convolution?

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  • $\begingroup$ $H(x)$ are Heaviside step functions right? $\endgroup$
    – tired
    Commented Mar 17, 2015 at 16:30
  • $\begingroup$ @tired: yes they are. let me try your answer and answer your question. thanks :) $\endgroup$ Commented Mar 17, 2015 at 16:31

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The expression $H(x)-H(x-\sigma)$ implies an range of integration from $0$ to $\sigma$ (Draw a picture for some specific values if you need to be convinced about that).

Setting $x=\frac{z-by}{a}$ in $F_X(x)$ it's a straightforward matter to see that we have to solve the integral: $$ F_Z(z)=\frac{\beta}{\sigma^2}\int_{\frac{z}{b}}^{\frac{z-a\sigma}{b}}\frac{z-by}{a}\left(\frac{y}{\sigma}\right)^{\beta-1}e^{-\left(\frac{y}{\sigma}\right)^{\beta}}dy $$

Now i would suggest to change variables $q=\left(\frac{y}{\sigma}\right)^{\beta}$. You should now be able to express the integral in terms of incomplete Gamma-Functions.

Therefore, using $dy=\frac{1}{\beta}q^{\frac{1}{\beta}-1}\sigma$ we have

$$ \frac{z}{a}\int_{\left(\frac{z\sigma}{b}\right)^{\beta}}^{\left(\frac{z-a\sigma}{b}\sigma\right)^{\beta}}e^{-q}-\frac{b}{a}\int_{\left(\frac{z\sigma}{b}\right)^{\beta}}^{\left(\frac{z-a\sigma}{b}\sigma\right)^{\beta}}q^{1/\beta}e^{-q}=\\ -\frac{z}{a}\left(e^{\left(\frac{z-a\sigma}{b}\right)^{\beta}}-e^{\left(\frac{z\sigma}{b}\right)^{\beta}}\right)-\frac{b}{a}\left[\gamma\left(\frac{1}{\beta}+1,\left(\frac{z-a\sigma}{b}\right)^{\beta}\right)-\gamma\left(\frac{1}{\beta}+1,\left(\frac{z\sigma}{b}\right)^{\beta}\right)\right] $$

Here $\gamma(a,x)$ denotes a lower incomplete Gamma-Function

Edit: Please note that this integral only exists for a specific range of parameters. I leave it to you to figure out this part rigorously!

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  • $\begingroup$ still finding for solution. can u help me with this please? i have more problems and hopefully i can solve them $\endgroup$ Commented Mar 17, 2015 at 16:46
  • $\begingroup$ Do you understand everything what i did so far? $\endgroup$
    – tired
    Commented Mar 17, 2015 at 16:50
  • $\begingroup$ yes i do. managed to derive on my own till this point $\endgroup$ Commented Mar 17, 2015 at 16:52
  • $\begingroup$ Following my proposition with the substitution there is really no further trick...Just some standard calculation. Everything i could do is just post the result, not give you any better advice... $\endgroup$
    – tired
    Commented Mar 17, 2015 at 16:54
  • $\begingroup$ okay. if u have worked it out, u can post it. else, i'll try to work it out :) $\endgroup$ Commented Mar 17, 2015 at 16:55

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