# $T$ exponential random variable, is $T - k$ a exponential random variable with same distribution as $T$ after a given time $k$?

Let $T$ be an exponential random variable with PMF $P_T(t) = \lambda e^{-\lambda t}$.

I am trying to show that after an arbitrary time $k$, the random variable $T - k$ is still exponential with parameter $\lambda$.

Here is my attempt.

$P(T \le t) = 1 - e^{-\lambda t}$, so $$P(T - k \le t | T > k) = \frac{P(T \le t + k)P(k < T \le t + k)}{P(T > k)}.$$

Computing these probabilities seperately, I get $$P(T \le t + k) = 1 - e^{-\lambda(t + k)}$$ $$P(k < T \le t + k) = P(T \le t + k) - P(T \le k) = e^{-\lambda k} - e^{-\lambda (t + k)} = e^{-\lambda k}(1 - e^{-\lambda t})$$ $$P(k > T) = 1 - P(T \le k) = 1 - (1 - e^{-\lambda k}) = e^{-\lambda k}$$

Therefore, plugging in we get, $$P(T - k \le t | T > k) = (1 - e^{-\lambda(t + k)})(1 - e^{-\lambda t}).$$

My expectation was that $P(T - k \le t| T > k) = 1 - e^{-\lambda t}$.

Can anyone explain where I went wrong? Or why my expectation is wrong?

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ah, just wrote this out on paper with arbitrary events A and B. And now I feel stupid. Thanks... heh – zrbecker Mar 19 '13 at 16:11

$$P(T-k\leq t\mid T>k)=\frac{P(T\leq t+k,\, T>k)}{P(T>k)}.$$ In other words, the numerator is $P(\{T\leq t+k\}\cap \{T>k\})$ which just corresponds to the second probability.