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$X$ is normally distributed with mean $\mu$ and variance $\sigma^2$ and d is a positive constant. How do I compute the following Expectation:$$E[(X-d) \textbf{1}_{(X>d)}]$$

This question is similar but doesn't give me an answer: Distribution of the indicator function of a Gaussian variable.

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HINT

$$ \mathbb{E}\left[ (X-d) \mathbb{I}_{(X>d)} \right] = \int_\mathbb{R} (x-d) \mathbb{I}_{(X>d)} f_X(x)dx = \int_d^\infty (x-d) f_X(x)dx, $$ where $f_X(x)$ is the pdf of $X$.

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