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数学代写|随机过程代写Stochastic Porcess代考|MATH11029

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数学代写|随机过程代写Stochastic Porcess代考|Definition

Let $X$ be a non-negative random variable with distribution function
$$F(x)=\operatorname{Pr}{X \leq x} .$$
The Laplace (Laplace-Stieltjes) transform $F^(s)$ of this distribution is defined, for $s \geq 0$, by $$F^(s)=\int_0^{\infty} \exp (-s x) d F(x)$$
We shall say that (3.1) also gives the “Laplace (Laplace-Stieltjes) transform of the random variable $X^{\prime \prime}$.
We have
$$F^(s)=E{\exp (-s X)}$$ and $$F^(0)=1 \text {. }$$
Suppose that $X$ is a continuous variate having density $f(x)=F^{\prime}(x)$. Then form (3.1),
$$F^(s)=\int_0^{\infty} \exp (-s x) f(x) d x$$ (this is the “ordinary” Laplace transform $L{f(x)} \equiv \bar{f}(s)$ of the density function $f(x)$ ). By a Laplace transform of a r.v. $X$, we shall mean the L.S.T. of the distribution function $F(\cdot)$ of $X$; this is equal to the ordinary L.T. of the density function of $X$ when this exists. We have $$F^(s)=\bar{f}(s) .$$
In case $X$ is an integral-valued random variable with distribution $p_k=\operatorname{Pr}{X=k}, k=0,1,2, \ldots$ and p.g.f. $P(s)=\sum_k p_k s^k$, we can stretch the language and define the L.T. of $X$ by
$$F^*(s)=E{\exp (-s X)}=P{\exp (-s)} .$$

数学代写|随机过程代写Stochastic Porcess代考|The Laplace Transform of the Distribution Function in Terms of that of the Density Function

Let $X$ be a continuous (and non-negative) r.v. having density function $f(x)$ and distribution function
$$\operatorname{Pr}{X \leq x}=F(x)=\int_0^x f(t) d t$$

The (ordinary) Laplace transform of the distribution function $F(x)$ is
$$L{F(x)}=\int_0^{\infty} \exp (-s x) F(x) d x$$
We have from A. 6 (Appendix)
\begin{aligned} \bar{F}(s) & =L{F(x)}=\int_0^{\infty} \exp (-s x)\left{\int_0^x f(t) d t\right} d x \ & =L\left{\int_0^x f(t) d t\right}=\bar{f}(s) / s . \end{aligned}
The relation can also be obtained by integrating by parts the relation (3.1). Thus we get
$$F^*(s)=\bar{f}(s)=s \bar{F}(s)$$
We note here that differentiation under the integral sign is valid for the L.T. given by (3.1), since the integrand is bounded and continuous.
Differentiating (3.1) with respect to $s$, we get
\begin{aligned} & \frac{d}{d s} F^(s)=-\int_0^{\infty} x \exp (-s x) d F(x) \ & \frac{d^2}{d s^2} F^(s)=(-1)^2 \int_0^{\infty} x^2 \exp (-s x) d F(x) \end{aligned}
and, in general, for $n=1,2, \ldots$
$$\frac{d^n}{d s^n} F^*(s)=(-1)^n \int_0^{\infty} x^n \exp (-s x) d F(x) .$$

随机过程代写

数学代写|随机过程代写Stochastic Porcess代考|Definition

$$F(x)=\operatorname{Pr}{X \leq x} .$$

$$F^(s)=E{\exp (-s X)}$$和$$F^(0)=1 \text {. }$$

$$F^(s)=\int_0^{\infty} \exp (-s x) f(x) d x$$(这是密度函数$f(x)$的“普通”拉普拉斯变换$L{f(x)} \equiv \bar{f}(s)$)。通过r.v. $X$的拉普拉斯变换，我们将表示$X$的分布函数$F(\cdot)$的拉普拉斯变换;当它存在时，它等于$X$的密度函数的普通L.T.。我们有$$F^(s)=\bar{f}(s) .$$

$$F^*(s)=E{\exp (-s X)}=P{\exp (-s)} .$$

数学代写|随机过程代写Stochastic Porcess代考|The Laplace Transform of the Distribution Function in Terms of that of the Density Function

$$\operatorname{Pr}{X \leq x}=F(x)=\int_0^x f(t) d t$$

$$L{F(x)}=\int_0^{\infty} \exp (-s x) F(x) d x$$

\begin{aligned} \bar{F}(s) & =L{F(x)}=\int_0^{\infty} \exp (-s x)\left{\int_0^x f(t) d t\right} d x \ & =L\left{\int_0^x f(t) d t\right}=\bar{f}(s) / s . \end{aligned}

$$F^(s)=\bar{f}(s)=s \bar{F}(s)$$ 这里我们注意到积分符号下的微分对于由(3.1)给出的l.t是有效的，因为被积函数是有界的和连续的。 (3.1)对$s$求导，得到 \begin{aligned} & \frac{d}{d s} F^(s)=-\int_0^{\infty} x \exp (-s x) d F(x) \ & \frac{d^2}{d s^2} F^(s)=(-1)^2 \int_0^{\infty} x^2 \exp (-s x) d F(x) \end{aligned} 一般来说，为 $n=1,2, \ldots$ $$\frac{d^n}{d s^n} F^(s)=(-1)^n \int_0^{\infty} x^n \exp (-s x) d F(x) .$$

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