I am reading a book titled Probability and Random Processes for Computer and Electrical Engineers and I came across a small tidbit which seems easy but I had a hard time showing why it was true. From page 140...
Given that $T \sim \text{exp}(\lambda)$, then $P(T \leq t + \Delta t | T \gt t) \approx \lambda \Delta t$. Now this seems like it should be easy to show so I just worked out the integral and I can't explain exactly why it is true...
$$ \begin{aligned} \int_x^{x + \Delta x} &= \lambda e^{-\lambda t} dx\\ &= \Big[ -e^{-\lambda x} \Big]^{t + \Delta t}_t \\ &= -e^{-\lambda(t + \Delta t)} (-) - e^{-\lambda(t)} \\ &= -e^{-\lambda t} e^{-\lambda \Delta t} + e^{-\lambda t} \\ &= e^{-\lambda t} (1 - e^{-\lambda \Delta t}) \\ \end{aligned} $$
but I cannot see how this should reduce to just $\lambda \Delta t$. Seems like I am missing something simple but I cannot see it. Have I setup the problem completely wrong?
Your caclulation seems a little messy... I won't point out all the mistakes you have made. I believe that my answer below will make things clear.
By definition, $$P(T\leq t+\Delta t|T>t)=\frac{P(t<T\leq t+\Delta t)}{P(T>t)},$$ so \begin{align*} P(T\leq t+\Delta t|T>t)&=\frac{\int_t^{t+\Delta t}\lambda e^{-\lambda x}\,dx}{\int_t^{\infty}\lambda e^{-\lambda x}\,dx}\\ &=\frac{-e^{-\lambda x}|_t^{t+\Delta t}}{-e^{-\lambda x}|_t^{\infty}}\\ &=\frac{e^{-\lambda t}-e^{-\lambda (t+\Delta t)}}{e^{-\lambda t}}\\ &=1-e^{-\lambda\Delta t}. \end{align*} Recall that $e^x\sim x+1$ for $x\to0$, so we have $$P(T\leq t+\Delta t|T>t)=1-e^{-\lambda\Delta t}\approx 1-(-\lambda\Delta t+1)=\lambda \Delta t,$$ here we consider $\Delta t$ very small.