What is the relationship between the Leibniz integral rule and the dominated convergence theorem?

We wish to prove $$ {displaystyle {frac {partial }{dx}}left(int _{a}^{b}f(x,t),dtright)=int _{a}^{b}{frac {partial }{partial x}}f(x,t),dt}. $$ We assume that $f:[c,d]times Tto mathbb{R}$ and $f(x, cdot)$ is integrable for each $X$. Further assume that there exists $gin L^1$ such that $|frac{partial f}{partial x} (x,t)|leq g(t)$ for all $x$ and $t$. Let $F(x)=int_a^b f(x,t) ,dt$.

To this end put $h_n(t)=frac{f(x_n, t)-f(x_0, t)}{x_n-x_0}$ where $x_nto x_0$. Then $ h_n(t)to frac{partial f}{partial x}(x_0, t) $ as $nto infty$. Further $$ |h_n(t)|leq sup_{xin[c, d]}|frac{partial f}{partial x} (x,t)|leq g(t) $$ Hence the dominated convergence theorem implies that $$ F'(x_0)=lim_{ntoinfty}frac{F(x_n)-F(x_0)}{x_n-x_0}=limint_a^b h_n(t),d t=int_a^b frac {partial }{partial x} f(x,t), dt $$ where in the last line we use the dominated convergence theorem.

First :

$frac{df(x)}{dtheta}=0$ is zero since $f$ doesn't depends on $theta$.

So your problem is rewritten like this :

There is two things to distinguish :

1-Dominated convergence theorem

Let a sequence of function $f_n$ : $X to Y$, converging pointwise to $f$ such that :

• There exists integrable $g$ such that for almost all $forall n in mathbb{N}$ , $x in X$ , $|f_n(x)|leq |g(x)|$

It follows that we can define the integral of $f$ :

$$ int f triangleqlim_infty int f_n = int lim_infty f_n = $$

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Now the Leibniz derivation rule is another theorem. (known as derivation under the sign) in which we use DCT to show it.

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2-Leibniz derivation rule

Taking a function $p:Theta times X to Y $ veryfing the follow conditions :

For all x $xin X$, la fonction $theta mapsto p(theta,x)$ is continous by part over $Theta$ and integrable over $Theta$;

$p$ admits partial derivative $frac{partial p}{partial x}$ defined on $Thetatimes X$;

For all $xin X$, the fonction $theta mapsto frac{partial p}{partial x}(theta,x)$ is continous by part over $Theta$;

For all $theta in Theta$, the fonction $xmapsto frac{partial p}{partial x}(theta,x)$ est continue sur $X$;

There exists $g:Thetatomathbb R_+$ continous by part and integrable, such that for all $xin X$ and for all $thetain Theta$, $$left|frac{partial p}{partial x}(theta,x)right|leq g(theta).$$

What you want to do is to bring the limit operation inside the integral sign. This cannot be done in general; one classic counterexample is that if

$$f_n(x)=begin{cases} n & x in [0,1/n] \ 0 & text{otherwise} end{cases}$$

then $$limlimits_{n to infty} int_0^1 f_n(x) dx = 1$$ but $$int_0^1 limlimits_{n to infty} f_n(x) dx = 0$$ The Leibniz rule for fixed limits can be justified by the use of the dominated convergence theorem, which is the most widely used method to prove that one can interchange limit and integral. It works fine in the setting of probability theory since it is based on the Lebesgue integral (which is how expectation is usually defined in probability theory).

As for the rest of your question, passing $frac{d}{dtheta}$ through $mathbb{E}_{p_theta}$ is definitely not valid, but $frac{d}{dtheta} mathbb{E}_{p_theta}[f(X)]=int frac{d}{dtheta} left [ f(x) p_theta(x) right ] dx$ is correct (assuming $p_theta$ is the density function of the distribution of $X$).