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Let $X$ follow an $\operatorname{Exponential}(\theta)$ and $Y = \min(X, 60).$ What is $\operatorname E_\theta[Y]$? Each value in Y is equivalent to X unless X is greater than 60 in those cases the Y value is 60.

My first attempt was to use the definition of expected value to evaluate the expectation in two different pieces.

$$\operatorname E_\theta[Y] = \sum_y p_x \cdot y =\sum_x p_y x + p[X \ge 60] \cdot 60$$

$$p[X \ge 60] = 1-\int_0^{60}\theta e^{-\theta x} \, dx = e^{-60\theta}$$

$$\operatorname E_\theta[Y] = \int_0^{60} \theta e^{-60\theta} + e^{-60\theta} 60 = 60e^{-60\theta}(\theta + 1)$$

This makes intuitive sense since the function of the expectation converges to $60$ as $\theta$ shrinks and the subsequent mean of $X$ increases. I am still not convinced this is the proper way to break down the expectation in the compound min.

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The expected value is \begin{align} & \int_0^{60} x f_X(x)\, dx + 60\Pr(Y=60) = \int_0^{60} xe^{-\theta x} (\theta\, dx) + 60\int_{60}^\infty e^{-\theta x} (\theta\,dx) \\[10pt] = {} & \frac 1 \theta \int_0^{60} (\theta x) e^{-\theta x} (\theta\,dx) + 60 \int_{60}^\infty e^{-\theta x} (\theta\,dx) \\[10pt] = {} & \frac 1 \theta \int_0^{60\theta} u e^{-u} \, du + 60 \int_{60\theta}^\infty e^{-u} \, du. \end{align}

The second integral comes to $e^{-60\theta}.$ The first can be done by parts: \begin{align} & \int u e^{-u} \, du = \int u\,dv = uv - \int v\, du = - u e^{-u} - \int-e^{-u} \, du \\[10pt] = {} & -60\theta e^{-60\theta} - \Big(e^{-60\theta} -1\Big) \end{align}

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