# Transformation of random variables vs shift of functions

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:

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:

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.
• 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!! Jan 17 at 13:25