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Let $Y$ and $Z$ two independent real valued random variables with values in $\mathbb{R}$ and let $f_Y(x)$ and $f_Z(x)$ their distributions.

Let $a, b \in \mathbb{R}$. How can the distibution $f_{aY + bZ}(x)$ of the linear combination $aY + bZ$ be simplified?

My ideas: If $a, b = 1$ I know that we have the folding formula $f_{Y + Z}(x) = f_Y * f_Z(x)$.

Futhermore, what about $f_{aY}(x)$? Does $f_{aY}(x)= f_Y(x/a)$ or $f_{aY}(x)= f_Y(ax)$ hold?

Can I combine this results in some way?

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2 Answers 2

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  1. Compute the pdf of $aY$. ($a\ne 0$)

CDF method:

$\begin{align} F_{aY}(x) &= P(aY\leq x) \\& = P(Y\leq \frac x a) \textbf 1_{a>0} + P(Y\geq \frac x a) \textbf 1_{a<0} \\ &= F(\frac x a) \textbf 1_{a>0} + (1-F(\frac x a)) \textbf 1_{a<0} \end{align}$

$f_{aY}(x) = \frac 1 a f(\frac x a) \textbf 1_{a>0} -\frac 1 a f(\frac x a) \textbf 1_{a<0} $

Finally, $$f_{aY}(x) = \frac 1{|a|} f_Y(\frac x a) $$

Expectation method:

Let $U=aY$.

For all h bounded, $$ \mathbb E(h(U))=\mathbb E(h(aY))=\int_{-\infty}^{\infty} h(aY)f_Y(y)dy $$

Now, perform the change of variable $x=ay$, $du=adx$. Note that the boundaries change when a is negative.

$$ \mathbb E(h(U))=\int_{-\infty}^{+\infty} h(x) \frac 1 a f_Y(\frac x a) dx \textbf 1_{a>0} + \int_{+\infty}^{-\infty} h(x) \frac 1 a f_Y(\frac x a) dx \textbf 1_{a<0} = \int_{-\infty}^{+\infty} h(u) \frac 1{|a|} f_Y(\frac u a)du $$

Yet $\forall \ h$ bounded, $ \mathbb E(h(U))=\int_{-\infty}^{\infty} h(x)f_U(x)dx$, defines an unique $f_U$. Therefore $$f_U(x) = \frac 1{|a|} f_Y(\frac x a) $$

  1. Now let's compute the pdf of the sum of the independent random variable.

The pdf of $U=aY$ is $f_U(x)=\frac 1{|a|} f_Y(\frac x a)$. Likewise, $V=bZ$ pdf is $\frac 1{|b|} f_V(\frac x b)$. Then since $Y$ and $Z$ are independent, the pdf of the sum is the product of convolution:

$$f_{U+V}(x) = (f_U * f_V)(x)$$

where * means convolution.

$$(f_U * f_V)(x) = \int_{-\infty}^{\infty}f_U(y)f_V(y-x)dy$$

Finally combining both results we get, $$f_{aY+bZ}(x) = \int_{-\infty}^{\infty}\frac 1{|a|} f_Y(\frac y a) \frac 1{|b|} f_Z( \frac{y-x}{b})dy $$

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  • $\begingroup$ Hi, thank you for the answer. If I understood you correctly then the only posibility to deduce the shape of $f_{aY}(x)$ is to work with CFD? Or can it also be done without it? The reason why I'm keen intersted in a proof without using CFD is to consider the case $f_{aY + bZ}(x)$. Here I can't apply the folding result to CDF. Do you see a posibility how to cope with this distribution? $\endgroup$
    – user267839
    Apr 23, 2018 at 20:57
  • $\begingroup$ I added how to use this result to compute the PDF or $aY+bZ$. The proof with CDF is the one that naturally came up to me. I will also add the method to directly deduce the PDF of $aY$ without using the CDF. $\endgroup$
    – Frostic
    Apr 23, 2018 at 21:07
  • $\begingroup$ Yes, thank you, again. $\endgroup$
    – user267839
    Apr 23, 2018 at 21:08
  • $\begingroup$ I added a second method (only using the PDF). The first method is more intuitive I think. I would recommend using the 2 method only if the 1 fails. $\endgroup$
    – Frostic
    Apr 23, 2018 at 21:24
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if $X = aY, f_X(x) = f_Y(\frac {y}{a})$

if $X = Y + Z, f_X(x) = \int_{-\infty}^{\infty} f_Y(y)f_Z(x-y)\ dy$

if $X = aY+bZ, f_{X}(x) = \int_{-\infty}^{\infty} f_Y(\frac {y}{a}) f_Z(\frac {x}{b}-\frac {y}{a})\ dy$

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  • $\begingroup$ The function you suggest as PDF does not integrate to $1$. $\endgroup$
    – Did
    Apr 23, 2018 at 20:30

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