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 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 $a&gt;0$. 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 $t_0&gt;0$ this corresponds to a shift to the right, and for $t_0&lt;0$ this corresponds to a shift to the left.