0
$\begingroup$

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:

enter image description here

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

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.