Assume the demand of a product $x$ is normally distributed with mean $\mu$ and standard deviation $\sigma$. If the order quantity of this product is $y$, then the profit will be $ax-b(y-x)$ for $y\ge x$ and $ay$ for $y<x$ where $a$ and $b$ are positive. How to choose $y$ to maximize profit?
I interpret this problem as the following: $X$~$N(\mu,\sigma)$, define $Z=aX-b(y-X)$ for $X\le y$ and $ay$ for $X>y$, and we need to choose a $y$ to maximize the expectation of $Z$. I think firstly we need to express the expectation of $Z$ explicitly, and so we need to compute CDF $F(z)=P(Z\le z)$ and then take derivative we will get PDF of $Z$ and then we can compute the expectation. Now, I am not sure what is CDF $F$ here.
Moreover, intuitively I guess we should choose $y$ that maximize $g(x)=xf(x)$($f(x)$ is the PDF of $X$) as the order quantity.
Thanks for any hint.
Fix $y$. Then $Z$ is a deterministic function of $X$, i.e. $Z=G(X)$ where $G(x) = ax-b(y-x)$ for $x<y$ and $G(x) = ay$ for $x>y$. Now computing the expectation of $Z$ is easy : $$\mathbb E[Z] = \mathbb E[G(X)] = \int_{\mathbb R} G(x)f_X(x)dx$$ where $f_X$ is the PDF of $X$. So $$\mathbb E[Z] = \int_{-\infty}^y[ax-b(y-x) ]f_X(x)dx + \int_{y}^{+\infty}ayf_X(x)dx.$$ We want to optimize in $y$ so we differentiate in respect to $y$. Using Leibniz's differentiation rule, $$\frac d{dy}\mathbb E[Z] = [ay-b(y-y)]f_X(y)\times 1 + \int_{-\infty}^y-bf_X(x)dx\ \ - ayf_X(y) + \int_{y}^\infty af_X(x)dx. = a(1-F_X(y)) - b F_X(y) = a-(a+b)F_X(y)$$
It becomes 0 when $F_X(y) = a/(a+b)$.