$Probability - Expectation calculation for a function.$

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$Probability - Expectation calculation for a function.$

A company manufactures metal poles. Suppose the length of a pole is a random variable X, with mean $\mu _X$ and probability density function $f_X(x)$ . Poles are cut to obtain an exact length 𝐿. If the initial length of the pole is less than 𝐿, the entire pole is lost. If it is greater than 𝐿, the pole will be cut down to 𝐿, and the section left over is lost. We are interested in the random variable 𝑌, defined as the length of each piece lost.
Sketch the graph of the function 𝑔 that maps the pole length 𝑥 to the lost length 𝑦, and so derive $\mu _Y$ = 𝔼(𝑌) as a function of $f_X(x)$ and $\mu _X$.
Suppose that 𝑋 follows a normal distribution with mean $\mu _X$ and variance $\sigma _X ^2$ . Show that there exists a value $\mu ^*$ of $\mu _X$ that minimizes $\mu _Y$.

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What you have doen for the fisrt part is correct and
\[\mu_Y=\mu_X -L \int_L^{\infty}f_X(x)dx.\]
If $X$ has a normal distribution with mean $\mu_X$ and variance $\sigma_X^2$, then
\[\mu_Y=\mu_X -L \int_L^{\infty} \frac{1}{\sigma_X\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{X-\mu_X}{\sigma_X})^2}dx=\mu_X -\frac{L}{\sigma_X\sqrt{2\pi}} \int_L^{\infty} e^{-\frac{1}{2}(\frac{X-\mu_X}{\sigma_X})^2}dx.\]
Consider the function
\[g(\mu_X)=\mu_X -\frac{L}{\sigma_X\sqrt{2\pi}} \int_L^{\infty} e^{-\frac{1}{2}(\frac{X-\mu_X}{\sigma_X})^2}dx, \mu_X \geq 0\]
Then $g$ is a continuous function on $[0,\infty)$ and
\[\lim_{\mu_X \rightarrow \infty}g(\mu_x)=+\infty.\]
Hence $g$ is a coercive function and must have a finite minimum point in $[0, \infty)$. So there is a $\mu_x$ that minimizes $g(\mu_X).$ See the Theorem in the scond page of the following notes: https://www.math.usm.edu/lambers/mat419/lecture4.pdf

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$Probability - Expectation calculation for a function.$

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