### 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Methods for specifying probabilistic measures

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## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Distribution function

Let $P=P(A)$ be a probabilistic measure (probability) defined on $\sigma$-algebra of Borel sets $\beta(R)$ of a number scale. Consider the probability space $(R, \beta(R), P)$. For the interval $A=(-\infty, x] \in \beta(R)$ define the function $F=F(x)$ as
$$F(x)=P(-\infty, x], x \in R$$
Theorem 1. The function (1) has the following properties:
F1. If $x_{1}<x_{2}$ then $F\left(x_{1}\right) \leq F\left(x_{2}\right)$ (i.e. $F=F(x)$ is a monotonically nondecreasing function);
F2. $F(-\infty)=\lim {x \downarrow-\infty} F(x)=0, \quad F(+\infty)=\lim {x \uparrow+\infty} F(x)=1$;
F3. $F(x)$ is a right-continuous function $(F(x+0)=F(x), x \in R)$, it has left limits at each point $x \in R$.

Proof. The property $\mathbf{F 1}$ is the corollary of the following. As $A_{x_{1}}=\left(-\infty, x_{1}\right] \subseteq\left(-\infty, x_{2}\right]=A_{x_{2}}$, then by the property of probability $P\left(A_{x_{1}}\right) \leq P\left(A_{x_{2}}\right)$.
The property $\mathbf{F} 2$ is the corollary of the following. From $x_{n} \downarrow-\infty$ it implies that $\left(-\infty, x_{n}\right] \downarrow \varnothing$, from $y_{n} \uparrow+\infty$ it implies that $\left(-\infty, y_{n}\right] \uparrow R$. We also use a continuity from below (property $\mathbf{P} 3^{\prime \prime}$ ) and continuity from above (property $\mathbf{P} 3^{\prime}$ ) of probability and monotonicity of the function $F(x)$.

It is not difficult to see that property $\mathbf{F} 3$ is also a consequence of the properties of continuity from below and from above.

Definition 1. The function $F=F(x)$ satisfying properties $\mathbf{F 1}, \mathbf{F} 2, \mathbf{F} 3$ is called the distribution function on the number line $R$.

Thus, by the Theorem 1 , the distribution function $F=F(x)$, defined by (1), corresponds to each probability function $P$ in the $\operatorname{space}(R, \beta(R))$. It turns out that the converse is also true.

Theorem 2. Let the function $F=F(x)$ be the distribution function on the number scale $R=(-\infty,+\infty)$.

Then on $(R, \beta(R))$ there is the only one probabilistic measure $P$ such that for any interval $-\infty \leq a<b<+\infty$ the following takes place:
$$P(a, b]=F(b)-F(a)$$
Proof. By the theorem on the extension of the probability, for constructing a probability space $(R, \beta(R), P)$, it suffices to specify the probability $P$ on the algebra $\mathcal{A}$ generated by intervals of the form $(a, b]$ (because $\sigma(\mathcal{A})=\beta(R)$ ). But we know that any element $A$ of algebra $\mathcal{A}$ can be written in the form of a finite sum of disjoint intervals of the form $(a, b]$ :
$$A=\sum_{i=1}^{n}\left(a_{i}, b_{i}\right], a_{i}<b_{i} .$$
$\left(a_{i}, b_{i}\right.$ may be infinite). Let’s by definition
$$P_{0}(A)=\sum_{i=1}^{n}\left[F\left(b_{i}\right)-F\left(a_{i}\right)\right] .$$

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Multidimensional distribution function

Let $P$ be a probability (probabilistic function), defined on $\left(R^{n}, \beta\left(R^{n}\right)\right)$. Let’s define the function of $n$ variables:

$$F_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)=P\left(\left(-\infty, x_{1}\right] \times \ldots \times\left(-\infty, x_{n}\right]\right)$$
or in a more compact form,
$$F_{n}(x)=P(-\infty, x]$$
where
$$x=\left(x_{1}, x_{2}, \ldots, x_{n}\right),(-\infty, x]=\left(-\infty, x_{1}\right] \times \ldots \times\left(-\infty, x_{n}\right]$$
We now introduce the difference operator $\Delta_{a_{i}, b_{j}}: R^{n} \rightarrow R\left(a_{i} \leq b_{i}\right), i=1,2, \ldots, n$, acting according to the formula:
$$\Delta_{a_{i}, b i} F_{n}\left(x_{1}, \ldots, x_{n}\right)=F_{n}\left(x_{1}, \ldots, x_{i-1}, b_{i}, x_{i+1}, \ldots, x_{n}\right)-F_{n}\left(x_{1}, \ldots, x_{i-1}, a_{i}, x_{i+1}, \ldots, x_{n}\right)$$
Calculations show that
$$\Delta_{a_{1}, b_{1}}, \Delta_{a_{n} b_{n}} F_{n}\left(x_{1}, \ldots, x_{n}\right)=P(a, b]$$
where $(a, b]=\left(a_{1}, b_{1}\right] \times \ldots \times\left(a_{n}, b_{n}\right]$.
We see from this relation that unlike the one-dimensional case, the probability $P(a, b]$, generally speaking, is not equal to the difference $F_{n}(b)-F_{n}(a)$. As $P(a, b] \geq 0$, then from the last relation (5) it follows that for any $a=\left(a_{1}, \ldots, a_{n}\right), b=\left(b_{1}, \ldots, b_{n}\right)$, $a_{i} \leq b_{i}, i=1,2, \ldots, n$, the property takes place
FF4. $\Delta_{a_{1}, b_{1}} \ldots \Delta_{a_{n}, b_{n}} F\left(x_{1}, \ldots, x_{n}\right) \geq 0$.
This property FF4 of the function $F_{n}(x)$ is called a property of non-negative definiteness.

Further, using the property of lower continuity of a probability function $P$, we obtain that the function $F_{n}(x)$ is a right continuous function with respect to the set of variables:
FF3. For $x^{(k)}=\left(x_{1}^{(k)}, \ldots, x_{n}^{(k)}\right) \downarrow x=\left(x_{1}, \ldots, x_{n}\right)$,
$$F_{n}\left(x^{(k)}\right) \downarrow F_{n}(x) .$$
In the same way, the following properties of the function a $F_{n}(x)$ are easily proved:
FF2. If $x \uparrow y=(+\infty, \ldots,+\infty)$, then $\lim {x \uparrow y} F{n}(x)=F_{n}(+\infty,+\infty, \ldots,+\infty)=1$;

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Space

Let the family of cylindrical sets in $R^{\infty}$ with «bases») $B \in \beta\left(R^{n}\right)$ be denoted by $J_{n}(B)$ :
$$J_{n}(B)=\left{x \in R^{\infty}:\left(x_{1}, x_{2}, \ldots, x_{n}\right) \in B\right}, \quad B \in \beta\left(R^{n}\right)$$
Let $P$ be a probabilistic measure on a measurable space $\left(R^{\infty}, \beta\left(R^{\infty}\right)\right)$. For any $n=1,2, \ldots$ denote by
$$P_{n}(B)=P\left(J_{n}(B)\right), \quad B \in \beta\left(R^{n}\right)$$
The sequence of probabilistic measures $P_{1}, P_{2}, \ldots$, defined on measurable spaces $(R, \beta(R)),\left(R^{2}, \beta\left(R^{2}\right)\right)$ (respectively), satisfies the consistency properties
$$P_{n+1}(B \times R)=P_{n}(B), \quad n=1,2, \ldots$$
It turns out that the converse assertion is also true (this assertion follows from the following theorem, given without proof, ([9], pp. 178-180)).

Theorem 6. (Kolmogorov’s theorem on the extension of a probability measure on $\left.\left(R^{\infty}, \beta\left(R^{\infty}\right)\right)\right)$. Let $P_{1}, P_{2}, \ldots$ be a sequence of probabilistic measures on $(R, \beta(R))$, $\left(R^{2}, \beta\left(R^{2}\right)\right), \ldots$ possessing the property of consistency (8).

Then there exists a unique probabilistic measure $P$ on $\left(R^{\infty}, \beta\left(R^{\infty}\right)\right)$ such that for each $n=1,2, \ldots$
$$P\left(J_{n}(B)\right)=P_{n}(B), \quad B \in \beta\left(R^{n}\right) .$$
Example 3. An example of a probability distribution on $\left(R^{\infty}, \beta\left(R^{\infty}\right)\right)$.
Let $F_{1}(x), F_{2}(x), \ldots-$ a sequence of one-dimensional distribution functions.
Let’s define the functions
$$G_{1}\left(x_{1}\right)=F_{1}\left(x_{1}\right), \quad G_{2}\left(x_{1}, x_{2}\right)=F_{1}\left(x_{1}\right) F_{2}\left(x_{2}\right), \ldots$$
and the corresponding probabilistic measures on $(R, \beta(R)),\left(R^{2}, \beta\left(R^{2}\right)\right), \ldots$ the theorem 6 denote by $P_{1}, P_{2}, \ldots$ (respectively). Then it follows from Theorem 6 that there exists a probabilistic measure $P$ in $\left(R^{\infty}, \beta\left(R^{\infty}\right)\right)$ such that
$$P\left{x \in R^{\infty}:\left(x_{1}, x_{2}, \ldots, x_{n}\right) \in B\right}=P_{n}(B), \quad B \in \beta\left(R^{n}\right)$$
and in particular
$$P\left{x \in R^{\infty}: x_{1} \leq a_{1}, x_{2} \leq a_{2} \ldots, x_{n} \leq a_{n}\right}=F_{1}\left(a_{1}\right) F_{2}\left(a_{2}\right) \cdot \ldots \cdot F_{n}\left(a_{n}\right)$$

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Distribution function

F(X)=磷(−∞,X],X∈R

F1。如果X1<X2然后F(X1)≤F(X2)（IEF=F(X)是单调非减函数）；
F2。F(−∞)=林X↓−∞F(X)=0,F(+∞)=林X↑+∞F(X)=1;
F3。F(X)是一个右连续函数(F(X+0)=F(X),X∈R)，它在每个点都有限制X∈R.

(一种一世,b一世可能是无限的）。让我们根据定义

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Multidimensional distribution function

Fn(X)=磷(−∞,X]

X=(X1,X2,…,Xn),(−∞,X]=(−∞,X1]×…×(−∞,Xn]

Δ一种一世,b一世Fn(X1,…,Xn)=Fn(X1,…,X一世−1,b一世,X一世+1,…,Xn)−Fn(X1,…,X一世−1,一种一世,X一世+1,…,Xn)

Δ一种1,b1,Δ一种nbnFn(X1,…,Xn)=磷(一种,b]

FF4。Δ一种1,b1…Δ一种n,bnF(X1,…,Xn)≥0.

FF3。为了X(ķ)=(X1(ķ),…,Xn(ķ))↓X=(X1,…,Xn),
Fn(X(ķ))↓Fn(X).

FF2。如果X↑是=(+∞,…,+∞), 那么 $\lim {x \uparrow y} F {n}(x)=F_{n}(+\infty,+\infty, \ldots,+\infty)=1$;

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Space

J_{n}(B)=\left{x \in R^{\infty}:\left(x_{1}, x_{2}, \ldots, x_{n}\right) \in B\right} , \quad B \in \beta\left(R^{n}\right)J_{n}(B)=\left{x \in R^{\infty}:\left(x_{1}, x_{2}, \ldots, x_{n}\right) \in B\right} , \quad B \in \beta\left(R^{n}\right)

G1(X1)=F1(X1),G2(X1,X2)=F1(X1)F2(X2),…

P\left{x \in R^{\infty}:\left(x_{1}, x_{2}, \ldots, x_{n}\right) \in B\right}=P_{n}(B ), \quad B \in \beta\left(R^{n}\right)P\left{x \in R^{\infty}:\left(x_{1}, x_{2}, \ldots, x_{n}\right) \in B\right}=P_{n}(B ), \quad B \in \beta\left(R^{n}\right)

P\left{x \in R^{\infty}: x_{1} \leq a_{1}, x_{2} \leq a_{2} \ldots, x_{n} \leq a_{n}\right }=F_{1}\left(a_{1}\right) F_{2}\left(a_{2}\right) \cdot \ldots \cdot F_{n}\left(a_{n}\righ

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