Expectation of minimum with exponential distributions

Question

Suppose $t_i$ is a random variable with exponential distribution. Let $M$ be a fixed real number. My instructor says $$E[min\{t_i, M\}]=E[t_i]-E[t_i]\times P(t_i >M).$$ I am a little confused. If this is indeed true, then $$E[min\{t_i, M\}]=E[t_i]\times P(t_i \leq M).$$ How should we think about it? Any thoughts or comments will be appreciated.

Answer 1

Let $T$ be an exponential distribution with parameter $\lambda$,\begin{align} E[\min\{ T, M\}] & = \int_0^\infty f(t) \min\{t,M \} \, dt \\ &= \int_0^M f(t) t \, dt + \int_M^\infty Mf(t) \, dt \\ &= \int_0^\infty f(t) t \, dt - \int_M^\infty f(t) t \, dt + \int_M^\infty Mf(t) \, dt\\ &= E[T] - \int_M^\infty f(t) (t-M) \, dt \\ &= E[T] - \int_0^\infty v\lambda \exp(-\lambda(v+M) ) \, dv \\ &= E[T] - \exp(-\lambda M)\int_0^\infty v\lambda \exp(-\lambda v ) \, dv \\ &= E[T] - \exp(-\lambda M)E[T] \\ &= E[T] - E[T] Pr(T >M)\\ \end{align}