Let $f$ and $g$ be two pdf’s, let $p ∈ (0, 1)$ and let $h(x) = pf(x) + (1 − p)g(x), \ x ∈ R.$
(a) Show that $h$ is a pdf.
(b) If $X$ has pdf $f$ and $Y$ has pdf $g$, what does the pdf $h$ describe?
(a): I know that a function $c$ is a pdf iff $c(x)\ge 0, \ \forall x\in{R}$ and
$$\int_{-\infty}^{\infty}c(x) \ dx=1.$$
So since it's given that $f$ and $g$ are pdf's, the above must hold for both of them. Now I have that
$$\int_{-\infty}^{\infty}h(x) \ dx=\int_{-\infty}^{\infty}pf(x)+(1-p)g(x) \ dx=p\int_{-\infty}^{\infty}f(x) \ dx+(1-p)\int_{-\infty}^{\infty}g(x) \ dx= \\ = p+(1-p)=1.$$
Also, clearly $h(x)\ge 0$, so we are done.
Is this all correct?
b): I'm extremely unsure of this one but I'll try.
If $f_X(x)$ happens with probability $p$ then it does not happen with prbability $1-p$, which is the same as $g_Y(x)$ happening. So I suppose that $h$ describes the probability of $f_X(x)$ and $g_Y(x)$ happening?
Good job for part $(a)$.
For part $(b)$,
Let $$Z= \begin{cases} X, & \text{with probability } p\\ Y, & \text{with probability } 1-p\end{cases}$$
By total law of probability, then \begin{align} P(Z \le z) &= pP(X \le z)+(1-p)P(Y \le z) \end{align}
If you differentiate the equation above with respect to $z$, we obtain $h$.