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Let $Y\sim\text{Exp}(1)$ be a random variable. I denote the random variable $X$ as $X=\min(Y,1)$. The task is to find the moment generating function of $X$.

By simply calculating the probability I managed to find the CDF of $X$ is: $$F_X(t) = \begin{cases} 1-e^{-t}, & 0 \le t <1\\ 1, & t \ge 1\\ 0, & \text{otherwise} \end{cases}$$

Here I got stuck. Since $X$ is not a continuous variable, it does not have a PDF, and without it I do not know how to calculate the moment generating function. ($X$ is also not discrete).

I will appreciate some help.

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The definition of the moment generating function does not require a PDF. $$\phi_X(t) := E[e^{tX}].$$

From here, it may be helpful to write $$e^{tX} = e^{tX} \mathbf{1}\{Y > 1\} + e^{tX} \mathbf{1}\{Y \le 1\}$$ and compute the expectation of the two terms separately.


Edit for more details:

$$E[e^{tX} \mathbf{1}\{Y > 1\}] = E[e^{tY} \mathbf{1}\{Y > 1\}] = \int_1^\infty e^{ty} f_Y(y) \, dy = \cdots.$$ $$E[e^{tX} \mathbf{1}\{Y \le 1\}] = e^t E[\mathbf{1}\{Y \le 1\}] = e^t P(Y \le 1) = \cdots.$$

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  • $\begingroup$ Could you show me how? $\endgroup$ – Gabi G Dec 10 '18 at 7:25
  • $\begingroup$ Thank you! But I think the integral needs to be from 0 to 1 $\endgroup$ – Gabi G Dec 10 '18 at 11:59
  • $\begingroup$ Thank you! But I think the integral needs to be from 0 to 1 $\endgroup$ – Gabi G Dec 10 '18 at 11:59

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