Two characteristic functions agree at $t=1$

Question

Suppose that $X_1$ and $X_2$ have the same distribution. Then, $$\phi_{t X_1} (1) = \phi_{X_1}(t) = \phi_{X_2}(t) = \phi_{t X_2} (1).$$

Why do $tX_1$ and $tX_2$ have the same distribution or how to show that $\phi_{t X_1} = \phi_{t X_2}$? How can we use the concept of characteristic function to write a proof for that?

Answer 1

For any $u\in\mathbb R$ $$\phi_{tX_1}(u)=\phi_{X_1}(tu)=\phi_{X_2}(tu)=\phi_{tX_2}(u),$$ so $tX_1$ and $tX_2$ have the same characteristic function and therefore the same distribution.

Answer 2

If $X_1$ and $X_2$ have the same distribution. so do $f(X_1)$ and $f(X_2)$ for any Borel measurable function $f$ on the real line. In particular this is true for the function $f(x)=tx$.

Proof: $P(f(X_1) \in A)=P(X_1 \in f^{-1} (A))=P(X_2 \in f^{-1} (A))=P(f(X_2) \in A)$ for any Borel set $A$.