Differentiating:$$\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty f(X_1,X_2,\ldots,X_n)\varphi_1(x_1,\theta)\cdots\varphi_n(x_n,\theta)\,dx_1 \cdots dx_n$$ with respect to $\theta$.

The result is given in one line, (the next one). I do not understand how this is. (Statistics proof) Anyway the result given being:

$$\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty f(X_1,X_2,\ldots,X_n) \sum_{i=1}^n \left(\frac{\partial}{\partial \theta}\varphi(x_i,\theta)\frac{1}{\varphi(x_i,\theta)}\right) \varphi_1(x_1,\theta)\cdots\varphi_n(x_n,\theta)\,dx_1\cdots dx_n$$

  • $\begingroup$ shouldn't the $\varphi$ inside the brackets be indiced with $i$, too? $\endgroup$
    – Max
    Jul 23, 2016 at 17:27
  • 1
    $\begingroup$ You switch from $X_i$ to $x_i$. You should know that mathematics is case sensitive. $X\neq x$. $\endgroup$
    – Asaf Karagila
    Jul 23, 2016 at 17:29
  • $\begingroup$ yeah it should. $\endgroup$ Jul 23, 2016 at 17:36

2 Answers 2


Let us rewrite the product rule as follows: $$(fg)'=f'g+g'f=\frac{f'}{f}fg+\frac{g'}{g}fg=\left(\frac{f'}{f}+\frac{g'}{g}\right)fg$$ Yours is just the generalization to $n$ factors, but is handled in the exact same way.


The first question is: Under what circumstances is $$ \frac \partial {\partial\theta} \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \bullet\bullet\bullet $$ the same as $$ \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \frac \partial {\partial\theta} \bullet\bullet\bullet \text{ ?} $$

The next question is: Why is $$ \frac \partial {\partial\theta} \varphi_1(x_1,\theta)\cdots\varphi_n(x_n,\theta) $$ the same thing as $$ \sum_{i=1}^n \left(\frac{\partial}{\partial \theta}\varphi(x_i,\theta)\frac{1}{\varphi(x_i,\theta)}\right) \varphi_1(x_1,\theta)\cdots\varphi_n(x_n,\theta) \text{ ?} $$

The answer to that last is the product rule, in this form: \begin{align} & \frac d {d\theta} (abcefg) \\[10pt] = {} & \left( \frac{da}{d\theta}\right) bcefg + a\left( \frac{db}{d\theta}\right) cefg + ab \left( \frac{dc}{d\theta}\right) efg + abc \left( \frac{de}{d\theta}\right) fg + \cdots \\[10pt] = {} & \left( \frac{da}{d\theta} \cdot \frac 1 a\right) abcefg + \left( \frac{db}{d\theta} \cdot \frac 1 b \right) abcefg + \left( \frac{dc}{d\theta}\cdot \frac 1 c \right) abcefg + \left( \frac{de}{d\theta}\cdot \frac 1 e \right) abcefg + \cdots. \end{align}


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