Como encontrar

Como posso resolver isso? Eu preciso de equações intermediárias. Talvez a resposta seja $-tf(x)$ .

d d t [\int \infty t x f (x) d x]

$\frac{d}{dt} \left [\int_t^\infty xf(x)\,dx \right ]$

$f(x)$ é a função de densidade de probabilidade.

Ou seja, $\lim\limits_{x \to \infty} f(x) = 0$ e $\lim\limits_{x \to \infty} F(x) = 1$

fonte: http://www.actuaries.jp/lib/collection/books/H22/H22A.pdf p.40

Tentando equações intermediárias abaixo:

d d t [\int \infty t x f (x) d x] = d d t [[x F (x)] \infty t - \int \infty t F (x) d x] ? ?

$\frac{d}{dt} \left [\int_t^\infty xf(x)\,dx \right ] = \frac{d}{dt} \left [\left [xF(x) \right ]_t^\infty - \int_t^\infty F(x)\,dx \right ]??$

d d t \int a t f (x) d x = - d d t \int t a f (x) d x = - d d t (F (t) - F (a)) = F' (t) = f (t)

$\frac{d}{dt} \int_t^a f(x)\,dx = -\frac{d}{dt} \int_a^t f(x)\,dx = -\frac{d}{dt} (F(t)-F(a))=F'(t)=f(t)$

— Hiroaki Machida
fonte

Você quer dizer

? Talvez

Ou você quer dizerddt[∫∞txf(x) dx] $\displaystyle \dfrac{d}{dt} \left [\int_t^\infty xf(x)~dx \right ]$

−tf(t). $-tf(t).$

? ddt[∫∞txf(x) dx1−F(t)] $\displaystyle \dfrac{d}{dt} \left [\dfrac{\int_t^\infty xf(x)~dx}{1-F(t)} \right ]$

— Henry

Use o teorema fundamental do cálculo

— Henry

Consider a primitive

G $G$ of

x↦xf(x) $x \mapsto xf(x)$ , then

∫∞txf(x)dx=G(∞)−G(t) $\int_t^\infty xf(x)dx=G(\infty)-G(t)$ is easy to derivate.

— Stéphane Laurent

Please add the self-study tag and read its tag wiki.

— Glen_b -Reinstate Monica

Se você está estudando para um exame, não é a solução ideal para você. As perguntas de auto-estudo têm como objetivo levar a pessoa que faz a pergunta a resolver o problema sozinha.

— Xian

Respostas:

Por definição, a derivada ( se existir ) é o limite do quociente de diferença

1 h (\int \infty t + h x f (x) d x - \int \infty t x f (x) d x) = - 1 h \int t + h t x f (x) d x

$\frac{1}{h}\left(\int_{t+h}^\infty x f(x) dx - \int_t^\infty x f(x) dx\right) = -\frac{1}{h}\int_t^{t+h} x f(x) dx$

como . $h\to 0$

Assumindo que é contínuo dentro de um intervalo para suficientemente pequeno , também será contínuo ao longo deste intervalo. Então o Teorema do Valor Médio afirma que existe algum entre e $f$ $[t, t+h)$ $h\gt 0$ $x f$ $h^{*}$ $0$ $h$ for which

- (t + h *) f (t + h *) = - 1 h \int t + h t x f (x) d x .

$-(t+h^{*})f(t+h^{*}) = -\frac{1}{h}\int_t^{t+h} x f(x) dx.$

As $h\to 0$ , necessarily $h^{*}\to 0$ , and the continuity of $f$ near $t$ then implies the left hand side has a limit equal to $-t f(t)$ .

(It is nice to see that this analysis requires no reasoning about the existence of the original improper integral $\int_t^\infty x f(x) dx$ .)

However, even when a distribution has a density $f$ , that density does not have to be continuous. At points of discontinuity, the difference quotient will have different left and right limits: the derivative does not exist there.

This is not a matter that can be dismissed as being some arcane mathematical "pathology" that practitioners can ignore. The PDFs of many common and useful distributions have points of discontinuity. For example, the Uniform $(a,b)$ distribution has discontinuous PDF at $a$ and $b$ ; a Gamma $(a,b)$ distribution has a discontinuous PDF at $0$ when $a\le 1$ (which includes the ubiquitous Exponential distribution and some of the $\chi^2$ distributions); and so on. Therefore, it is important not to assert, without careful qualifications, that the answer is merely $-t f(t)$ : that would be a mistake.

— whuber
fonte

A very small addendum: There are cases where the integral is differentiable even when

f(x) $f(x)$ is not continuous. Let

f(x)=0 $f(x)=0$ for

x≤0 $x\leq 0$ and

f(x)=1 $f(x)=1$ for

0<x<1 $0<x<1$ and

f(x)=0 $f(x)=0$ for

x≥2 $x\geq 2$ . Then near 0,

F(x)=x2/2 $F(x)=x^2/2$ for

x≥0 $x\geq 0$ and 0 for

x<0 $x<0$ , which is perfectly differentiable at

x=0 $x=0$ .

— Alex R.

@Alex Near

0+ $0^{+}$ ,

F(x)=x $F(x)=x$ , not

x2/2 $x^2/2$ . Consider the Fundamental Theorem of Calculus.

— whuber

Sorry for the confusion! I define

F(x):=∫x−∞tf(t)dt $F(x) := \int_{-\infty}^x tf(t)dt$ .

— Alex R.

@Alex Your integrand

tf(t) $tf(t)$ is continuous near zero, so I fail to see what kind of example you are presenting or what it shows.

— whuber

Great derivation (+1) - it might be worth nothing that this result is a case of Leibniz integral rule.

— Reinstate Monica

Solved...

$\displaystyle \dfrac{d}{dt} \left [\int_t^\infty xf(x)~dx \right ]$ $= \displaystyle \dfrac{d}{dt} \left [G(\infty)-G(t) \right ]$ $= \displaystyle \dfrac{d}{dt} \left [G(\infty) \right ] - \dfrac{d}{dt} \left [G(t) \right ]$ $= \displaystyle 0-tf(t)$

Thank you all!!!

— Hiroaki Machida
fonte

What is

G(t) $G(t)$ function? Why the derivative of

G(∞) $G(\infty)$ is 0?

— Vladislavs Dovgalecs