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I am looking for the derivative of the expected value of the truncated normal distribution with respect to one of the upper limits. It looks like this: $$f=\frac{1}{\sqrt{\vert T \vert} (2\pi)^{3/2}} \int_{-\infty}^{x_3=a_3} dx_3 \int_{-\infty}^{x_2=a_2} dx_2 \int_{-\infty}^{x_1=a_1} dx_1 \begin{bmatrix}{x_1\\x_2\\x_3}\end{bmatrix} exp\bigg(-0.5\begin{bmatrix}{x_1 x_2 x_3}\end{bmatrix} T^{-1} \begin{bmatrix}{x_1\\x_2\\x_3}\end{bmatrix} \bigg)$$\ $$\frac{\partial f}{\partial a_2}?$$

I have tried using the Fundamental Theorem of Calculus but I am a bit stuck what to do with the other integrals and elements of the vectors.

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The answer is \begin{equation} \frac{\partial f}{\partial a_2}=\frac{1}{\sqrt{\vert T \vert} (2\pi)^{3/2}} \int_{-\infty}^{x_3=a_3} dx_3 \int_{-\infty}^{x_1=a_1} dx_1 \begin{bmatrix}{x_1\\a_2\\x_3}\end{bmatrix} exp\bigg(-0.5\begin{bmatrix}{x_1 a_2 x_3}\end{bmatrix} T^{-1} \begin{bmatrix}{x_1\\a_2\\x_3}\end{bmatrix} \bigg). \end{equation}

Consider the antiderivative $F(x)$ of the function \begin{equation} F(x) = \int f(x) dx \end{equation} or \begin{equation} \frac{d F(x)}{dx} = f(x), \end{equation} so definite integral is the difference of antiderivatives \begin{equation} \int\limits_a^b f(x)dx = F(b) - F(a), \end{equation} i.e. the derivative is \begin{equation} \frac{\partial}{\partial b} \int\limits_a^b f(x)dx = F^{\prime}(b) = f(b). \end{equation}

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  • $\begingroup$ The exp() here is just a scalar function. So actually you have three integrals of scalar functions (one integral for the each component of the vector f). $\endgroup$
    – S.G.
    Commented Mar 4, 2021 at 12:51

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