A PAM signal is $x(t) = \sum_n a_n h(t-nT)$, where $\{a_l\} =\{\pm 1\}$ is a sequence of random binary bits, $h(t)$ is a pulse shape, and $T$ is the bit period. The FFT of $x(t)$ is: $$X(f) = \int_{-\infty}^{\infty} \sum_n a_n h(t-nT) e^{-j2 \pi ft} dt$$ $$ = \sum_n a_n \int_{-\infty}^{\infty} h(t-nT)e^{-j2 \pi ft} dt$$ $$ = \sum_n a_n H(f) e^{-j2 \pi fnT}$$ $$ = H(f)A(e^{-j2 \pi fT})$$
where $A(e^{-j2 \pi fT})$is the DTFT of $a_n = a(nT)$ with a frequency scaling factor of $T$.
I want to compute the PSD of $x(t)$. Can someone explain why the PSD turns out to be:
$$\frac{E\{a_n^2\}}{T}|H(f)|^2$$
I don't understand why the scaling factor $1/T$?