I have problem deriving the expectation of random variable from a delayed differential equation. Any comment or advice for getting the result would be much appreciated!
The distribution of random variable $q \in [0, \infty]$ satifies
$$qf(q) = A [F(q)-F(q-B)]$$
where $F(q)$ is CDF, $f(q)$ is PDF and $A$ and $B$ are some constants.
If we integrate both sides over the domain, we get
$$\begin{align} \int_0^{\infty} qf(q) dq &= A \int_0^{\infty} [F(q)-F(q-B)] dq \\ &=A \left[\int_0^{\infty} [F(q)-1] dq + \int_0^{\infty} [1-F(q-B)] dq\right]\\ &=A \left[-\int_0^{\infty} qf(q) dq + \int_0^{\infty} [1-F(q-B)] dq\right] \\ \int_0^{\infty} qf(q) dq &= \frac{A}{1+A} \left[\int_0^{\infty} [1-F(q-B)] dq\right] \end{align}$$ Then changing of variable $x = q-B$, we have $q = x+B$ and $dq = dx$. So $$\begin{align} \int_0^{\infty} qf(q) dq &= \frac{A}{1+A} \left[\int_{-B}^{\infty} [1-F(x)] dx\right] \end{align}$$ The answer says that as $F(X)=0$ for $x\leq 0$, then $$\begin{align} \int_0^{\infty} qf(q) dq &= AB \end{align}$$ But I am so confused. I was thinking the following derivation: $$\begin{align} \int_{-B}^{\infty} [1-F(x)] dx &= \int_{0}^{\infty} [1-F(x)] dx + \int_{-B}^{0} 1 dx \\ &=\int_{0}^{\infty} [1-F(x)] dx + B \\ &=\int_{0}^{\infty} x f(x) dx + B \end{align}$$ And this is where I am. Is this right and how can I get the answer?
What you've done so far looks good to me, and you're almost there! If you put $\ E=\int_\limits{0}^\infty qf(q)\,dq\ $, the expectation you're trying to calculate, then your equations $$ \int_\limits{0}^\infty qf(q)\,dq= \frac{A}{1+A} \left[\int_{-B}^{\infty} [1-F(x)] dx\right] $$ and $$ \int_{-B}^{\infty} [1-F(x)] dx=\int_{0}^{\infty} x f(x) dx + B $$ give you $$ E=\frac{A}{1+A}[E+B]\ . $$ I'm sure you're capable of solving this for $\ E\ $.