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I have this following case (please refer to attached pic below) where a particle is resting on the ground and it needs a minimum amount of force (Fmin) to reach from one level to the next level.

But if at any level it receives F which is more than Fmin, then the excess force can be carried forward to the next levels, and hence even a force less than Fmin will be enough for it to go to the next level.

The force F follows a two-sided truncated gaussian distribution, and has only positive values.

So the probability that the object moves from 0 to 1 is P0=P(F>Fmin) where F = two-sided truncated gaussian distribution with specified mean, and std. dev.

Now suppose, when the particle was at level 0, it received force F0>Fmin, reached level 1, and now the probability to go from 1 to 2
P1 = P(F>(Fmin-excess F it received earlier)
= P(F>Fmin-(F0-Fmin))
= P(F>2Fmin-F0)

At level 1, it had received force F1, so now probability from level 2 to 3 is
P2 = P(F>3Fmin-(F0+F1))

and again from level 3 to target is
P3 = P(F>4Fmin-(F0+F1+F2))

So, here the probability at any level is dependent on the F values it received in the previous stages.

I guess the final probability can be written as
P = P0*P1*P2*P3

Is there any probability model for this type of multi-stage motion? How can I model this problem. Any help or suggestions.

If my question is not clear, please let me know. Thanks.

Edit 1:
If it receives more than Fmin at level 0, it will definitely reach level 1, and can even reach level 2, or even the target level depending on the F magnitude.

Suppose at level 0 it receives only Fmin, so it reaches level 1, but at level 1 it receives less than Fmin, In this case it falls to the below level that is level 0. Suppose this thing happens at level 2, then the particle falls back to level 1 initially, and if it doesn't get Fmin even at level 1, it finally goes to level 0.

So the particle cannot stay in any level for an extended time, except level 0 where it is waiting for Fmin. At other levels, it if doesn't have Fmin it falls back to the below level.

Edit 2:
The force distribution is updated to a two-sided truncated Gaussian distribution to have only positive forces. and to reflect the possibility that the particle may not even reach the target eventually.

The particle receives a force at all levels, irrespective of the force magnitude it received at previous levels.

For clarity I must add that if the particle falls down even by 1 level, all the accumulated excesses are now gone, and it will again need at least Fmin to again start over from its present level.

Simulated this motion in Matlab. So I am just adding an output image which demonstrates one of the many possible up-and-down movements of the particle until the target level is reached. Hope it will make the question easier to understand.

Edit 3:
Added the updated image of the particle motion. Hope this helps.

enter image description here

enter image description here

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  • $\begingroup$ What happens when it doesn't receive enough force? $\endgroup$
    – Ian
    Jun 17, 2016 at 4:49
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    $\begingroup$ How often does it receive force? Is it a once only impetus, or does it receive a boost each level? $\endgroup$ Jun 17, 2016 at 5:04
  • $\begingroup$ @GrahamKemp It receives a force at every level, only the magnitude it receives at every level differs. However at each level, it needs to have at least Fmin to move on to next level. $\endgroup$ Jun 17, 2016 at 5:09
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    $\begingroup$ It stays at the same level, but does it "accumulate" force, or if it fails to go to the next level is it now "back where it started"? Also, what happens if at one stage it receives enough positive force to jump two levels but then receives enough negative force that it goes back down a level? (Note that negative force is necessarily possible by the Gaussian assumption). $\endgroup$
    – Ian
    Jun 17, 2016 at 19:22
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    $\begingroup$ why is $P0 = P(F > Fmin)$ and not $P0 = P(F0 > Fmin)$ ? Is there a concept of time ? If it receives anough force to jump several levels, does it still get extra force for all the intermediate levels ? Does it receive force even after it falls down ? Can we have different $F1$s if it goes through the 1st level several times ? If it goes to the target is it stuck there forever ? Why is the final probability not $1$ ? $\endgroup$
    – mercio
    Jun 19, 2016 at 15:52

1 Answer 1

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Your process has random samples $X_t$ at times $t=0,1,2,\ldots$ from a continuous state space $[0,4]$. The state $x = 4$ is absorbing and reachable from every transient state, so this is an absorbing Markov chain.

I'm going to restate the specific rules of the random process for the sake of definiteness. I believe my interpretation is consistent with the Comments taken as a whole, but the graph illustrating a possible outcome has some peculiarities that would be worth clarification by the OP. Since the main question posed is "How can I model this [multi-stage motion]", it seems to me these details are less important than the approach in general.

Let $\nu$ be a continuous probability measure on the reals, such as a standard unit normal, having a positive probability $\mathrm{Pr}(V\ge 1)$ of values greater than $1$.

We assume initially that $X_0 = 0$. Subsequently, for each positive integer $t$ we sample a value of $V_t$ from the designated continuous distribution. Then $X_t\in [0,4]$ is determined from this $V_t$ and the prior state $X_{t-1}$ according to the following rules:

  • If $X_{t-1} = 4$, then $X_t = 4$ no matter what value $V_t$ has, so $X_t = 4$ is an absorbing state.
  • If $X_{t-1} \lt 4$ and $X_{t-1} + V_t \ge \lfloor X_{t-1} \rfloor + 1$, then $X_t = \min(X_{t-1} + V_t , 4)$.
  • If $X_{t-1} \lt 4$ and $X_{t-1} + V_t \lt \lfloor X_{t-1} \rfloor + 1$, then $X_t = \max(\lfloor X_{t-1} \rfloor - 1 , 0)$.

By standard arguments the random process $X_t$ remains in $[0,4)$ until eventually hitting the absorbing state $X_t = 4$ with probability one.

As @Ian comments, we can derive the probability measure for each $X_t$ . These are combinations of continuous densities on $[0,4]$ and discrete masses on $\{0,1,2,3,4\}$. To this end let's define probability density functions $\omega_t$ for each nonnegative integer time $t$, from which the random variables $X_t$ are respectively sampled.

We are given that the initial probability "density" is the Dirac delta function:

$$ \omega_0 = \delta $$

a generalized function which concentrates a unit mass at the origin $x=0$.

The probability measure $\omega_1$ then consists of discrete mass at the endpoints $x=0,4$ together with a continuous measure with support on $[1,4]$. In terms of the underlying probability measure $\nu$ from which $V_t$ is sampled, we can write the probability measure from which $X_t$ is sampled:

$$ \omega_1(x) = \left( \int_{-\infty}^1 \nu(s) ds \right) \delta(x) + \mathbf{1}_{[1,4]}(x) \nu(x) + \left( \int_{4}^\infty \nu(s) ds \right) \delta(x-4)$$

where $\mathbf{1}_S$ is the indicator function for subset $S$, taking value $1$ there and zero elsewhere.

For $t \gt 1$ the probability measure $\omega_t$ will have discrete positive masses at each node $x=0,1,2$ and $4$ and a continuous positive density whose support is $[1,4]$. I'm working on trying to present the linear mapping from $\omega_t$ to $\omega_{t+1}$ in a reasonably concise form.

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  • $\begingroup$ One thing that is unclear in the description of the force levels is what happens when the force received at a level is enough to "lift" it two or more levels at once. Taking the wording of the Question at face value, the amount of "retained" force would be sufficient to raise it to the next level whatever the (positive) increment received at just one level up. However it is not entirely clear if a force increment should be received both at one-level up and at two-levels up, etc. in such a case. $\endgroup$
    – hardmath
    Jun 19, 2016 at 16:22
  • $\begingroup$ Reply to your comment: Yes, you are right that when force received is high enough, it can cover two levels or more at once. I had also mentioned this is Edit 1: "If it receives more than Fmin at level 0, it will definitely reach level 1, and can even reach level 2, or even the target level depending on the F magnitude." To answer your next question, yes, it will still receive a force in this case at all the higher levels 1-3 irrespective of how big a force it received at level 0. $\endgroup$ Jun 20, 2016 at 4:07
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    $\begingroup$ Thanks for editing the Question to clarify how the failure of advancing to the next level affects the particle and its accumulated "force". As my answer assumes the particle's advance toward the "final target" was failed at such a point, it needs revision. However if the particle can return to $\mathbf{F_0}$ an arbitrary number of times, success is ultimately a probability one outcome, regardless of the use of a full or a truncated Gaussian distribution (so long as the jumps $\mathbf{F_{min}}$ are possible). $\endgroup$
    – hardmath
    Jun 20, 2016 at 13:38
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    $\begingroup$ Oh yes, you are right about the probability being 1 as long as the particle can receive Fmin. Thanks for correcting me. Will edit this in the question. I am also adding an image of one of the possible movements of the particle until it reaches the target level. Hope it will make the question easier to understand. $\endgroup$ Jun 20, 2016 at 14:44
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    $\begingroup$ I think you can also probably proceed directly, using a state space like $[0,4]$. I see two problems: 1. you really only care about the "projection", i.e. when the state is $x \in [0,4]$ you really only care what $\lfloor x \rfloor$ is. 2. the transition probability kernel of the process is weird. It has a continuous part and a considerable discrete part as well. This is likely to make it difficult to do the usual renewal techniques to write down an equation to solve your problem. $\endgroup$
    – Ian
    Jun 20, 2016 at 15:32

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