# CDF and conditional probability

The cdf of $T$ is given by $F_T(t)=1-e^{-\lambda t}$ for $x>0$. Show that $P(T>t+s\mid T>s)=P(T>t)$. Find the MGF of $T$.

I don't really know what's going on. Do I substitute $t+s$ into the $t$ in the original cdf?

For the MGF do I first find the pdf? If so do I differentiate wrt $t$ or $x$?

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1. Find $\mathbb{P}\left(T>t\right)$ using $F_T(t)$.
2. Use definition of conditional probability: $$\mathbb{P}\left(T>t+s\mid T>s\right) = \frac{\mathbb{P}\left(T>t+s,T>s\right)}{\mathbb{P}\left(T>s\right)} \stackrel{\text{assuming }s>0}{=} \frac{\mathbb{P}\left(T>t+s\right)}{\mathbb{P}\left(T>s\right)}$$
$P(T>t)=1-(1-e^{-\lambda t})=e^{-\lambda t}$. If that is right, the rest of the proof is easy. Also please see my edit about the MGF. –  bbr4in Feb 8 '13 at 19:31
@user52187 To find pdf, recall $f_T(t) = \frac{\mathrm{d}}{\mathrm{d}t} F_T(t)$. Then compute the expectation defining the moment generating function $\mathcal{M}_T(x) = \mathbb{E}\left( \exp(T x) \right)$. –  Sasha Feb 8 '13 at 21:52
I seem to be dealing with a partial integral here. I am sure I am doing something wrong. Please take a look at this: $\int_0^{\inf} \lambda e^{-\lambda t} e^{Tx} dx$ –  bbr4in Feb 10 '13 at 22:34