Convergence in distribution and shifted pdf

Question

Suppose $X_n \xrightarrow{d} X$ and the pdf $f_{Y_n}$ of $Y_n$ is defined by $f_{Y_n} (x) = f_{X_n}(1+x)$. Is it true that $Y_n \xrightarrow{d} X$?

By Portmanteau, $$\lim \mu_{X_n}(A) = \mu_X(A), \quad \text{for all $\mu$-continuity sets $A \in \mathcal \Sigma$},$$

if and only if that

$$E[f(X_n)] \to E[f(X)], \quad \text{for all bounded, continuous function $f$.}$$

Answer 1

Hint: $$ F_{Y_{n}}(a)=\int_{-\infty}^{a}f_{Y_{n}}(y)dy=\int_{-\infty}^{a}f_{X_{n}}(1+y)dy=\int_{-\infty}^{1+a}f_{X_{n}}(x)dx=F_{X_{n}}(1+a). $$

Answer 2

$f_{Y_n} (x) = f_{X_n}(1+x)$ means that the distribution of $Y_n$ is the same as distribution of $X_n-1$. There is no reason for $X_n-1$ to converge in distribution to $X$. It clearly converges to $X-1$.