Spacecraft Space Jump

 

When a spacecraft is operating in the space, the control system of the spacecraft fails, and the spacecraft cannot jump to space normally, but can only set one distance in advance, then, the system performs a completely random jump from this distance. The problem now is that the spacecraft wants to return to the solar system. Assume that the radius of the solar system is $ r$. The distance between the flying ship and the sun is $ r$ when a fault occurs. Here, $ r> r$. At every moment, the ship can know its distance from the solar system.

Proof: No matter what Skip strategy is adopted, the probability that a spacecraft returns to the solar system is less than $ R/r$. However, if $ \ Epsilon> 0 $ is used, an appropriate strategy can be adopted, the probability of a spacecraft returning to the solar system is greater than $ (R-\ epsilon)/r$, that is, $ R/r$ is the optimal probability. What is this optimal policy?

This interesting question comes from Williams's teaching material "probability with Martingales". Like other articles in this series, this article demonstrates that the abstract Yang theory has a variety of applications.

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Preparations for analysis

Theorem:Set $ S $ to $ \ mathbb {r} ^ 3 $ to a sphere with a center of $ A $ and a radius of $ r$, $ x $ is the random point evenly distributed on $ S $, then the expectation value of the reciprocal distance between $ x $ and the origin $ o $

\ [E \ frac {1 }{| x |}=\ left \{\ begin {array }{ L} 1/A & R <, \ 1/R & R \ geq. \ end {array} \ right. \]

Here, $ A $ and $ | x | $ represent the distance between $ A $ and $ x $ and the origin $ o $ respectively.

This is actually a high school physical knowledge we are all familiar with: Suppose there is a uniform distribution of charge with a total amount of 1 on the Sphere $ S $, the potential at the origin $ o $ (ignoring physical constants) is the expected $ e \ frac {1 }{| x |}$.

We know that on the even charged shell $ S $ and inside $ S $, the potential is constant, equal to $1/r$; the potential at the external point of the shell $ S $ is equal to the reciprocal of the distance from the point to the ball center.

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Prerequisites

The general condition is expected to be defined by the derivative of the map-Nikodym, which is sometimes not convenient to use. For example, if $ X and Y $ are two random variables, if event $ \ {x = x \} $ is a zero probability event, but $ x = x $ is known, the expected value $ f (x) = E [Y | x = x] $ is obviously probabilistic, that is, the $ x $ information is given, the expected value of $ y $ is $ f (x) $. Therefore, $ E [Y | x] = f (x) $ is displayed. But how can we explain that this intuitive Condition Expectation is consistent with the abstract Condition Expectation? This is the following "freezing Lemma ".

Theorem:Set $ (\ Omega, \ mathcal {f}, p) $ to a probability space, $ \ mathcal {g} \ subset \ mathcal {f} $ is a child $ \ Sigma-$ algebra with random variables $ X: \ Omega \ rightarrow e $ about $ \ mathcal {g} $ measurable, random variable $ Y: \ Omega \ rightarrow F $ is independent from $ \ mathcal {g} $ (thus $ X, Y $ is independent). Here $ E and F $ are two measurable state spaces. If $ F: E \ times f \ rightarrow \ mathbb {r} $ is a non-negative Measurable Function (or a Bounded Measurable Function ),

\ [E [F (x, y) | \ mathcal {g}] = E [F (x, y)]. \]

Or write it in a more intuitive form: \ [E [F (x, y) | \ mathcal {g}] = E [F (x, y) | x = x]. \]

The proof of the theorem is the standard method starting from the explicit function $ every A \ times every B $.

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Model Creation

1. The initial time is set to 0, and the solar system is centered around the origin and the sphere with a radius of $ r$. the initial position of the spacecraft is $ (R, 0, 0) $.

2. Set $ \{\ overrightarrow {u_n }\}_{ n \ geq 1 }1 to define a probability space $ (\ Omega, \ mathcal {f }, \ mathbf {p}) $ represents a group of independent and uniformly distributed random vectors on the unit sphere, which represent the random direction of each space jump of a spacecraft. Set $ \ mathcal {f }_n = \ sigma (\ overrightarrow {U_1}, \ ldots, \ overrightarrow {u_n }) $ and $ \ mathcal {f} _ 0 = (\ Omega, \ emptyset) $.

3. Set the distance to $ N $ space jump to $ l_n (n \ geq1) $, because the distance between the first $ N $ space jump is determined based on the previous information, so $ l_n $ about $ \ mathcal {f }_{ n-1} $ is measurable.

4. So if the coordinates of the ship after the $ N $ space jump are $ \ overrightarrow {x_n} $, then \ [\ overrightarrow {x_n} = \ overrightarrow {X _ {n-1} + l_n \ overrightarrow {u_n }. \ quad n =, \ ldots \] where $ \ overrightarrow {x_0} = (R,) $ is the initial position of the spacecraft.

5. Set $ T $ to the first time the spacecraft returns to the solar system: \ [T = \ INF \ {n: |\overrightarrow {x_n} \ in B (0, R) \\},\] Then the value range of $ T $ is $ \ mathbb {n ^ +} \ cup \ {+ \ infty \} $. We need to estimate the probability of event $ \ {T <+ \ infty \} $, which is the probability that the spacecraft can return to the solar system within a limited time.

Now let's start to study the movements of the ship.

In fact, it is enough to figure out what will happen during the first hop, because the next hop is just a "recurring scene" of the first hop ". So what will happen after the first jump?

Set $ R_N $ to the distance between the spacecraft and the solar system after the $ N $ jump, $ R_0 = r$, and apply the previous theorem 1.

\ [E \ left [\ frac {1} {R_1} \ right] =\left \ {\ begin {array} {L} 1/R & L_1 <R, \ 1/l_1 & L_1 \ geq R. \ end {array} \ right. \]

In particular, no matter how you set $ L_1 $, there will be $ E [1/R_1] \ Leq 1/r$. That is to say, in the expected sense, the derivative of the distance between the spacecraft and the solar system after it jumps is decreasing!

We can immediately apply this phenomenon to every next hop: at $ n-1 $, the distance between the spacecraft and the solar system is given $ R _ {n-1} $, no matter how you set the Jump Distance $ l_n $ for the first $ N, there is always

\ [E \ left [\ frac {1} {R_N} \ left | \ right. R _ {n-1} = A \ right] \ Leq \ frac {1} {}. \]

That is

\ [E \ left [\ frac {1} {R_N} \ left | \ right. \ mathcal {f }_{ n-1} \ right] \ Leq \ frac {1} {R _ {n-1 }}. \]

In a word

Theorem 1:$ \ {1/R_N \} $ about $ \ {\ mathcal {f} _ n \} $ constitutes an upper-raised question (irrelevant to the policy ). In particular, if the distance between the current spacecraft and the solar system is not more than the distance, that is, for $ n \ geq1 $ l_n \ Leq R _ {n-1} $, then, $ \ {1/R_N \} $ is a raised sign.

Proof: you only need to describe that each $1/R_N $ is a random variable that can be accumulated. Because $1/R_N $ is a non-negative random variable, $ E [\ frac {1} {R_N} | \ mathcal {f }_{ n-1}] $ is defined and less than or equal to $1/R _ {n-1} $. Sum up $ N $.

Theorem 1 is the most critical step to solve the entire problem. When it comes to it, the sky is wide and the sky is wide. If it doesn't exist, it will be hard to implement. From this, we can immediately export an interesting observation: because non-negative upper-Yang always converges almost everywhere, therefore, the conclusion of Theorem 1 contains $ \ lim \ limits _ {n \ To \ infty} R_N (\ Omega) $, which exists almost everywhere (the limit can be positive infinity ), there are two possibilities: $ \ lim \ limits _ {n \ To \ infty} R_N (\ Omega) = + \ infty $ or $ \ lim \ limits _ {n \ To \ infty} R_N (\ Omega) = A <+ \ infty $. So the ship either flies to infinity, that is, gets lost in the depths of the universe, or is "banned" in a limited area.

Now we can prove that:

Theorem 2:Regardless of the policy adopted by the spacecraft, the probability of returning to the solar system is strictly less than $ R/r$.

Prove that only a very basic knowledge is used:

If $ Z_N = 1/R_N $ is set, then $ \ {Z_N \} $ is a non-negative upper gradient, therefore, $ Z _ \ infty = \ lim \ limits _ {n \ To \ infty} Z_N $ exists almost everywhere. Consider the non-negative top-yang sequence $ \ {z _ {T \ wedge n }\}$ obtained by the truncation of the period $ T $. This top-yang sequence converges almost everywhere, where \ [\ lim _ {n \ To \ infty} Z _ {T \ wedge n }=\ left \ {\ begin {array} {L} \ lim _ {n \ to \ infty} Z_N & t = \ infty, \ z_t & T <\ infty. \ end {array} \ right. \]

On the one hand, \ [E [\ lim _ {n \ To \ infty} Z _ {T \ wedge n}] \ Leq \ varliminf _ according to the Fatou theorem of the non-negative Product Function Column _ {n \ To \ infty} e [Z _ {T \ wedge n}] \ Leq E [z_0] =\ frac {1} {r }. \]

On the other hand

\ [E [\ lim _ {n \ To \ infty} Z _ {T \ wedge n}] = E [z_t1 _ {\{ T <\ infty \}] + E [\ lim _ {n \ To \ infty} z_n1 _ {\{ T = \ infty \}] \ geq E [z_t 1 _ {\{ T <\ infty \ }] \ geq \ frac {1} {r} \ mathbf {p} (T <\ infty ). \]

The last non-equal sign is $ z_t \ geq 1/r$. Based on these two inequalities, we get $ \ mathbf {p} (T <\ infty) \ Leq R/r$, that is, the probability that a spacecraft eventually returns to the solar system under any policy is no greater than $ R/r$.

To prove that the probability is strictly less than $ R/r$, we only need to prove that the last non-equal sign of the formula above is strictly true: \ [E [z_t 1 _ {\{ T <\ infty \}}] >\frac {1} {r} \ mathbf {p} (T <\ infty ). \]

Of course, assume that $ \ {T <\ infty \} $ has a positive probability (if the return probability is 0, it is certainly less than $ R/r$, and there is no evidence ).

Therefore, you only need to prove that $ R_N <r$ is almost everywhere on Event $ \ {T <\ infty \} $.

We can prove a more refined conclusion:

\ [A_n :=\ {R_N = r \ Text {or} R_N = 0 \}. \]

We need to prove that each $ n \ geq0 $, $ a_n $ is a zero probability event.

Sum up $ N $: $ n = 0 $ R_0 = r> r$. The conclusion is true. If $ k <N $ has $ \ mathbf {p} (A_k) = 0 $, for $ k = N $, we use

\ [E [1 _ {a_n} | \ mathcal {f }_{ n-1}] =\ mathbf {p} (a_n | \ mathcal {f }_{ n-1 }) = 0. \ quad \ Text {. e. $ \ Omega \ In \ mathcal {f_n }$ }. \]

Here $ \ mathbf {p} (a_n | \ mathcal {f }_{ n-1 }) $ almost everywhere 0 is the use of inductive assumptions $ \ mathbf {p} (A _ {n-1}) = 0 $ and obvious geometric facts: after the zero test set $ A _ {n-1} $ is not considered, for $ \ Omega \ notin a _ {n-1} $, the probability of a hop occurring in the center of the solar system or on a sphere $ r$ from the center of the solar system is 0. Therefore

\ [\ Mathbf {p} (a_n) = E [E [1 _ {a_n} | \ mathcal {f }_{ n-1}] = 0, \]

The conclusion is also true for $ N $.

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Now we are moving our attention to the event "The ship cannot return to the solar system. As we have already said before, there are only two possibilities for a spacecraft to move. They are lost in an infinite distance or restricted in a limited area. Therefore, if the spacecraft cannot return to the solar system, it will either fly to an infinite distance, either in a limited area outside the solar system. We want to know how to determine which of the two situations will happen?

For example, consider a obviously unreasonable strategy: the distance between the first $ n and the next hop is always set to $1/2 ^ N $, in this strategy, the spacecraft can never fly out of a space with a radius of 1, so this strategy should be avoided.

The problem with this bad strategy is that the sum of hop distances converges. In fact, if we add a divergence limit to the jump distance, we can avoid the condition that the spacecraft is spinning around somewhere outside the solar system. This is the theorem below:

Theorem 3:Set \ [E :=\{\ omega: \ t (\ Omega) =\infty, \ sum _ {L = 1} ^ \ infty l_n (\ Omega) = \ infty \}, \]

Then we have

\ [\ Lim _ {n \ To \ infty} R_N = + \ infty, \ quad \ Text {for A. E. $ \ Omega \ In e $.} \]

The conclusion of Theorem 3 is that if a spacecraft cannot return to the solar system ($ T = \ infty $ ), the jump distance is still divergent ($ \ sum \ limits _ {n} l_n = \ infty $), so the spacecraft is almost always flying to infinity.

The proof of Theorem 3 is also interesting. It combines knowledge of both geometric and probability.

Set $ \ theta_n $ to the angle between $ \ overrightarrow {X _ {n-1 }}$ and $ \ overrightarrow {u_n} $, the link $ \ overrightarrow {x_n} =\ overrightarrow {X _ {n-1} + l_n \ overrightarrow {u_n} $ and the cosine formula of the triangle are available.

\ [R_N ^ 2 = R _ {n-1} ^ 2 + l_n ^ 2 + 2R _ {n-1} l_n \ cos \ theta_n. \]

For $ B _n =\{\ cos \ theta_n \ geq 1/2 \}$, we have

\ [R_N ^ 2 \ geq R _ {n-1} ^ 2 + l_n ^ 2 + R _ {n-1} l_n \ geq (R _ {n-1} + \ frac {l_n }{ 2 }) ^ 2. \]

\ [R_N \ geq R _ {n-1} + \ frac {l_n} {2}, \ quad \ Omega \ In e \ cap B _n. \]

Of course

\ [R_N \ geq R _ {n-1} + \ frac {l_n} {2} \ Text {I. O .} \ quad \ Omega \ In e \ cap \ {B _n \ Text {I. O .} \}. \]

Because the coordinate components of random points evenly distributed on the sphere are also evenly distributed, \ [\ mathbf {p} (B _n) = \ mathbf {p} [\ overrightarrow {u_n} \ In \ {(x, y, z) \ In \ mathbb {r} ^ 3: \ Z \ geq \ frac {1} {2} \}] =\ frac {1} {4}> 0. \]

Event column $ \ {B _n \} $ is independent and $ \ sum \ limits_n \ mathbf {p} (B _n) = \ infty $, according to Borel-cantelli's second theorem, $ \ mathbf {p} (\ {B _n \ Text {I. O .} \}) = 1 $, so $ e \ cap \ {B _n \ Text {I. O .} \} $ and $ e $ only have one zero test set, which proves

\ [R_N \ geq R _ {n-1} + \ frac {l_n} {2} \ Text {I. O .} \ quad \ Text {for. e. $ \ Omega \ In e $ }. \]

Note that $ \ lim \ limits _ {n \ To \ infty} R_N $ always exists and $ \ sum \ limits_n l_n (\ Omega) = \ infty (\ Omega \ in E) $, which immediately means

\ [\ Lim _ {n \ To \ infty} R_N = + \ infty \ quad \ Text {for A. E. $ \ Omega \ In e $}. \]

This proves the theorem.

Note: Why is the event column $ \ {B _n \} $ independent? First, you must note that

\ [\ Mathbf {p} (B _n | \ mathcal {f }_{ n-1}) =\ frac {1} 4 {}. \]

This is because the location information of the given spacecraft at $ n-1 $ is, the angle distribution between the next hop direction $ \ overrightarrow {u_n} $ and $ \ overrightarrow {X _ {n-1} $ can be calculated. The key here is that the conditional probability is a constant, so $ \ mathbf {p} (B _n | \ mathcal {f }_{ n-1}) = \ mathbf {p} (B _n) $, that is, the probability of event $ B _n $ does not depend on $ \ mathcal {f }_{ n-1} $, which proves that $ \ {B _n \} $ is an independent event column.

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Now we know that the probability of a spacecraft returning to the solar system is always less than $ R/r$. We also know that, as long as the strategy is right, the spacecraft can be switched around in the same place. The following question is: what is the best strategy?

Theorem 4:Define the following skip strategy: when preparing the $ N $ jump, if the spacecraft is already in the solar system, set $ l_n to 0 $, otherwise, $ l_n = R _ {n-1}-R + \ Epsilon $. Here, $0 <\ Epsilon <r$. In this hop strategy, the probability that a spacecraft returns to the solar system is greater than $ (R-\ epsilon)/r$.

Note that there is always $ l_n <R _ {n-1} $ in this policy. Therefore, $ \ {Z_N \}=\{ 1/R_N \}$ is actually a gradient. In addition, because $ R_N \ geq R-\ Epsilon $ is always available, $ Z_N \ leq1/(R-\ epsilon) $, that is, $ \ {Z_N \} $ is controlled by the constant $1/(R-\ epsilon) $.

The following proof is just a repetition of Theorem 2:

According to the control convergence theorem

\ [E [\ lim _ {n \ To \ infty} Z _ {T \ wedge n}] = \ lim _ {n \ To \ infty} e [Z _ {T \ wedge n}] = E [z_0] = \ frac {1} {r }. \]

On the other hand

\ [E [\ lim _ {n \ To \ infty} Z _ {T \ wedge n}] = E [z_t1 _ {\{ T <\ infty \}] + E [\ lim _ {n \ To \ infty} z_n1 _ {\{ T = \ infty \}]. \]

At this time, we should note that there is always $ l_n \ geq \ Epsilon $ in $ \ {T = \ infty \} $. Therefore, according to the conclusion in Theorem 3, the spacecraft will almost inevitably fly to infinity, that is

\ [\ Lim _ {n \ To \ infty} Z_N = 0 \ quad \ Omega \ In \ {T = \ infty \}, \]

So

\ [E [\ lim _ {n \ To \ infty} Z _ {T \ wedge n}] = E [z_t1 _ {\{ T <\ infty \}] \ leq \ frac {1} {r-\ Epsilon} \ mathbf {p} (T <\ infty ). \]

The two formulas show that $ \ mathbf {p} (T <\ infty) \ geq (R-\ epsilon)/r$.

Spacecraft Space Jump