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Suppose that for standard brownian motion $\{B(t): t\geq 0\}$, we define the stopping time $T_a = \inf \{ t \geq 0: B(t) = a\}$. Using the pdf of $T_a$ only: $$f_{T_a}(t) = \frac{a}{\sqrt{2\pi}} e^{-\frac{a^2}{2t} t^{-\frac{3}{2}}}\hbox{, }t> 0$$ how can we show that $T_a = \frac{1}{c}T_{a\sqrt{c}}$ in distribution?

bilbo
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If $X$ is a random variable with density $p$, then $$\mathbb{P}(\tfrac{1}{c} X \in A) = \int 1_A(\tfrac{1}{c} x) p(x) \, dx = c \int 1_A(y) p(cy) \, dy.$$ Since you know that the random variable $X=T_{a \sqrt{c}}$ has density $$p(t)=\frac{a \sqrt{c}}{\sqrt{2\pi}} \exp \left(- c \frac{a^2}{2t} \right) t^{-3/2},$$ it follows that \begin{align*} \mathbb{P}(\tfrac{1}{c} T_{a \sqrt{c}} \in A) &= c \int 1_A(y) p(cy) \, dy \\ &= c \int 1_A(y) \frac{a \sqrt{c}}{\sqrt{2\pi}} \exp \left(- c \frac{a^2}{2(cy)}\right) (cy)^{-3/2} \, dy \\ &= \int 1_A(y) \frac{a}{\sqrt{2\pi}} \exp \left(-\frac{a^2}{2y} \right) y^{-3/2} \, dy \\ &= \mathbb{P}(T_a \in A).\end{align*} Since this holds for any measurable set $A$, it follows that $\tfrac{1}{c} T_{a \sqrt{c}} = T_a$ in distribution.

saz
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