I am a beginner to random variables and I am understanding the concept of the transformations of a random variable. Consider a random variable $X$ to be Gaussian distributed with $a_x = 1.6$ and $\sigma_x = 0.4$. The random variable now undergoes a transformation $Y=X-1.6$. We know that the relationship between $Y$ and $X$ is given by: \begin{align} f_Y(y) = f_X(x)\left\vert \frac{dx}{xy} \right\vert \end{align}.$$ Y=X - 1.6 \implies X = Y+1.6 \\ \therefore \left\vert\frac{dx}{dy} \right\vert = 1 $$ $$ \textit{Thus,} \ \ f_Y(y) = f_X\left( Y+1.6 \right) \\ \implies f_Y(y) = \frac{1}{\sqrt{2\pi\sigma_x^2}}exp\left(-\frac{(Y-(a_x - 1.6))^2}{2\sigma_x^2}\right)$$ with $a_y = a_x - 1.6 = 0.0$ and $\sigma_x = \sigma_y = 0.4$ Graphically, it looks like this:

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However,another concept that is bugging me is the transformation of independent variables in signals and systems, like the shift operation. Using this concept, $f(t-1.6)$ would be a delayed version of the function $f(t)$ as follows:

enter image description here

I know this may look stupid, but can someone please help me clear out this confusion?

Thank you so much.


I think you've actually answered your own question by observing that the pdf of the random variable (RV) $Y=X-a$ is given by $f_Y(y)=f_X(y+a)$, which corresponds to a left shift for $a>0$. This is also immediately clear from the fact that by subtracting $a$ from $X$, the mean of the transformed RV is also shifted to $m_Y=m_X-a$.

Shifting a general function $f(t)$ is just the same, and you actually did exactly that with the pdf of the original RV to obtain the pdf of the transformed RV. For a shifted function $f(t-t_0)$, the origin (i.e., the value $f(0)$) is shifted to the value $t=t_0$. So for $t_0>0$ this corresponds to a shift to the right, and for $t_0<0$ this corresponds to a shift to the left.

  • $\begingroup$ Oh... I see. My bad, I should have checked for $f_Y(y)$ instead of $f_X(x)$. And I think you mean to say subtracting $a$ from $X$ instead of $Y$? Anyways, thank you so much!! $\endgroup$ Jan 17 at 13:25
  • $\begingroup$ @NishanthRao: Sure, and yes, I meant what you wrote, editing ... $\endgroup$
    – Matt L.
    Jan 17 at 13:26

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