In the probability course on Youtube at 6:23, the professor said that:
$$f_{X+b}(x) = f_{X}(x-b)$$
Why is this the case? Could someone offer a proof?
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Consider the CDF of $X + b$:
$$ F_{X+b}(x) = \Pr\{X + b \leq x\} = \Pr\{X \leq x - b\} = F_X(x - b)$$
where $F_{X+b}, F_X$ are the CDF of $X+b$ and $X$ respectively. Differentiate both sides with respect to $x$, we have
$$ \begin{align} \frac {\partial} {\partial x} F_{X+b}(x) &= \frac {\partial} {\partial x} F_X(x - b) \\ \Rightarrow f_{X+b}(x) &= \frac {\partial} {\partial (x - b)}F_X(x - b) \times \frac {\partial (x - b)} {\partial x} \\ &= f_X(x-b)\end{align}$$
where it is well-known that differentiating the CDF will obtain the corresponding pdf
BGM
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Hi BGM, the chain rule part where $F_{x}(x-b)$ differentiated by x seems to be a typo of differentiated by (x-b)? But thanks a lot! – SayMyNameHeisenberg Mar 07 '20 at 15:02