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Probability density and rifle shooting

  1. Jan 5, 2015 #1
    1. The problem statement, all variables and given/known data
    diagram-png.54451.png
    A rifle shooter aims at a target at a distance D, but has an accuracy probability density
    ρ(φ)=1/(2Φ) φ∈(-Φ,Φ)
    where φ is the angle achieved and is bounded by the small angle Φ


    Part A
    find the probability density for where the bullet strikes the target , ρ(x) . The target has a width of 2d . H denotes the event where the bullet hits the target whereas M denotes that when it misses. Calculate and depict the probability of hitting
    P(H;Φ) as a function of Φ for fixed d and D with d=Dtanθ

    Part B
    as well as the intrinsic accuracy, the shooter can also set his sights, described by the angle ψ, with respect to the original zero angle . Find the modified probability P(H;Φ;ψ).

    Part C
    if ψ is also randomly distributed with probability density
    ρ(ψ)=1/(2Ψ) ψ∈(-Ψ,Ψ)

    Part D
    By considering N shots between resighting, explain how to test whether a shooter is limited by the accuracy or by accuracy of sighting

    2. The attempt at a solution

    Part A
    ρ(φ)dφ=ρ(x)dx ; x=Dtanφ ; dx=Dsec(φ)2

    Therefore, ρ(x)=(1/2Φ)D/(D2+x2)
    For θ ≤ Φ
    P(H;Φ) = 1
    P(H;Φ)=1-P(M;Φ)=1-2*(θΦρ(φ)dφ)
    = θ/Φ

    Part B

    For Φ>ψ,
    θ<abs(Φ-ψ) :P(H;Φ;ψ)=1-P(M;Φ;ψ)=1-(θΦ+-(Φ-ψ))ρ(φ)dφ
    abs(Φ-ψ)<θ<(Φ+ψ) : P(H;Φ;ψ)=1-P(M;Φ;ψ)=1-θΦρ(φ)dφ
    +ψ)<θ :P(H;Φ;ψ)=1

    For Φ<ψ,
    θ<abs(Φ-ψ) :P(H;Φ;ψ)=0
    abs(Φ-ψ)<θ<(Φ+ψ) : P(H;Φ;ψ)=1-P(M;Φ;ψ)=1-θΦρ(φ)dφ
    +ψ)<θ :P(H;Φ;ψ)=1

    Part C

    This is where I start to get stuck, as I have no idea how I could go about combining the two probability density functions to give a distribution function. Looking up wiki, I notice that it might involve something named ' convolution', which I have no knowledge of it whatsoever, and I can't seem to find any alternative ways.

    Part D

    This is where I get almost completely clueless as I don't really understand what I'm being asked to do. I suspect it might involve something like comparing the size of two probability distributions limited by different factors (accuracy or accuracy of sighting) . Any help would be greatly appreciated.
     
  2. jcsd
  3. Jan 5, 2015 #2

    Stephen Tashi

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    I don't understand the notation "[itex] 1/(2\phi)[/itex]". I assume the density is [itex] P_\phi(x) = 1/2 [/itex] for [itex] x [/itex] in [itex] [-\phi,\phi] [/itex] and zero elsewhere.

    Yes, to find the density of the sum of two independent random variables you calculate the convolution of their densities.

    The density function [itex] f(x) [/itex] of the random variable [itex] \phi + \psi [/itex] must tell us (roughly speaking) the probability that [itex] \phi + \psi [/itex] is in a small interval around [itex] x [/itex]. To compute this probability, we must consider all the ways that [itex] \phi + \psi [/itex] can add up to [itex] x [/itex]. This amounts to considering the probability that of the event [itex] ( \phi = x - h [/itex] and [itex] \psi = h ) [/itex] for all possible values of [itex] h [/itex]. When [itex] \psi [/itex] and [itex] \phi [/itex] are independent random variables, that probability (roughly speaking) is the product of the densities [itex] p_\phi(x-h) p_\psi(h) [/itex]. To add up all these probabilities in the case of discrete random variables, we would do a summation over all values of [itex] h [/itex] where the densities have a non-zero probability. For continuous random variables we do an integration with respect to [itex] h [/itex]. This calculation is called a "convolution".

    This integration for a convolution is often written with the infinite limits as [itex] \int_{-\infty}^{\infty} p_\phi(x-h) p_\psi(h) dh [/itex] with the understanding that the densities are defined on all real numbers (so they are defined to be zero on impossible values). In a practical problem, you often have to determine finite limits for the integrals since the ordinary calculation of [itex] \int_{-\infty}^{\infty} ....dh [/itex] in calculus doesn't consider that the integrand has special conditions that set it zero outside some finite interval.

    I don't know if your course deals with statistical tests. The answer may have something to do with comparing the mean location of a batch of shots made with one sighting to a the mean location of a batch of shots taken with a different sighting.
     
  4. Jan 5, 2015 #3

    Ray Vickson

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    What, exactly, is meant by "setting the sights" in part B? Although your writeup does not at all made this issue clear, it looks like the effect of choosing ##\psi## is, perhaps, to made the shooting angle come out uniform over the interval ##(-\Phi + \psi, \Phi + \psi)##. Is that the case? Also, it sounds as though the shooter sets the sights deliberately and manually, so that ##\psi## is some fixed number. It then seems strange to have it be random, as in part D; but I guess an exercise in probability does not have to make practical sense, just mathematical sense.

    After these issues are clarified I will be better able to offer comments.

    BTW: if the new accuracy really is of the form ##\phi + \psi## with both terms random, the "accuracy" ##\gamma = \phi + \psi## after setting the sights will be more spread out (from ##-\Phi-\Psi## to ##\Phi+\Psi##), but will also be more closely concentrated near ##\gamma = 0##, so you would need to cook up some type of measure the say whether one situation is better or worse than the other. That is precisely the issue you mentioned in your final paragraph. There would be an element of subjectivity in that, because two fair-minded people might have opposite opinions about the comparisons. Typical methods used in such situations would be (i) comparisons of standard deviations of the distributions; (ii) comparision of hitting probabilities; (iii) comparision of "central probabilities" ##P( -a < X < a)## for some fixed ##a < d##. The answers from those three methods need not agree, since they are emphasizing different aspects.

    As for "convolution": there are numerous explanatory web pages at various levels of sophistication. Basically, the probability density of a sum of random variables is the convolution of their individual probability densities.

    Note added in edit: after reading Stephen Tashi's response, I realize that looking at a sum ##\phi+\psi## as a summed random variable may not be appropriate. If the effect of a given setting is unknown, but remains fixed throughout the multiple shots, then every shot is affected by the exact same (unknown) value of ##\psi## each time. Looking at the "sum" would only be appropriate if the sights were re-set between shots, so that the setting from one shot would not affect the settings for future shots. The scenario needs to be clarified before you could hope to analyze it further.
     
    Last edited: Jan 5, 2015
  5. Jan 5, 2015 #4

    Ray Vickson

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    Last edited: Jan 5, 2015
  6. Jan 5, 2015 #5
    I'm not too sure either. My interpretation is that ψ is just a chosen angle and the range of angles achieved is I ψ±φ I with respect to the horizon.
    If I haven't misunderstood this sentence, I believe that's the case. ψ is a fixed quantity throughout the N shots. But then I'm uncertain to which other probability setting is to be compared.
     
  7. Jan 5, 2015 #6

    haruspex

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    That doesn't look right. Do you mean
    For θ > Φ, P(H;Φ) = 1
    For θ ≤ Φ
    etc?
    Yes, that's how I read it, except that it should be Φ, not φ, and the range is ψ±Φ. No need for the ||. But your solution doesn't seem to have enough cases. E.g. (ψ+Φ > θ and ψ-Φ < -θ) is different from (ψ+Φ > θ and ψ-Φ > -θ).

    Edit: you may be puzzled by my quibble over the form of phi. On my laptop they looked rather different; on my ipad they look almost the same.
     
    Last edited: Jan 6, 2015
  8. Jan 6, 2015 #7

    Stephen Tashi

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    In military parlance there is "aiming error" [itex] \psi [/itex] and "round to round error" [itex] \varphi [/itex]. When you aim a weapon at a target, you make some "aiming error". Keeping the same aim, you discharge several rounds. We assume the discharge of the rounds doesn't disturb your aim. The rounds fall with various errors in a distribution centered on where you aimed, rather than the actual place you should have aimed.
     
  9. Jan 6, 2015 #8
    Yes I did mess up.
    P(H;Φ) = 1 when θ > Φ
    For θ ≤ Φ
    P(H;Φ)=1-P(M;Φ)=1-2*(θ∫Φρ(φ)dφ)
    = θ/Φ

    Yup thx for correcting it. Should there be eight different cases in total ?
     
  10. Jan 6, 2015 #9
    I think that's in agreement with my understanding on the angle ψ .. is it ?
     
  11. Jan 6, 2015 #10

    Stephen Tashi

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    Yes, for one shot. Part D inquires about taking multiple shots from the same aiming line.
     
  12. Jan 6, 2015 #11

    haruspex

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    i count six.
     
  13. Jan 6, 2015 #12
    In the counting process I keep ψ>0 to avoid overcounting the no. of cases
    Φ>ψ:

    1. ψ+Φ>θ ∩ ψ-Φ<-θ
    2. ψ+Φ>θ ∩ ψ-Φ>-θ
    3. ψ+Φ<θ ∩ ψ-Φ>-θ
    ψ+Φ<θ ∩ ψ-Φ<-θ (impossible with ψ>0 :ψ+Φ<θ⇒-(ψ+Φ)>-θ⇒-(ψ+Φ)>-θ>ψ-Φ⇒-ψ>ψ)

    Φ<ψ:

    ψ+Φ>θ ∩ ψ-Φ<-θ (impossible with Φ<ψ and ψ>0 , as 0<ψ-Φ)
    4. ψ+Φ>θ ∩ ψ-Φ>-θ
    5. ψ+Φ<θ ∩ ψ-Φ>-θ
    ψ+Φ<θ ∩ ψ-Φ<-θ (impossible with ψ>0 , the same as the above.)

    that's how I come up with 5...
     
  14. Jan 6, 2015 #13

    haruspex

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    Ok, but with that constraint I only count 4. Writing ##A = \Psi+\Phi##, ##B=\Psi-\Phi##, there's ##A>B>\theta##, ##A>\theta>B>-\theta##, ##A>\theta>-\theta>B##, ##\theta>A>B>-\theta##. Does that miss any?
     
  15. Jan 6, 2015 #14
    In A>B>θ the shooter would have no chance at hitting at all.

    and I'm not too sure whether I'm overcounting if I include 2 different scenarios for A>θ>B>-θ and θ>A>B>-θ , where B could be positive or negative.

    Edit : It wouldn't matter if B is positive or negative in θ>A>B>-θ as the hitting probability will just be 1 regardless. However, in A>θ>B>-θ, the probabities are different
     
    Last edited: Jan 6, 2015
  16. Jan 6, 2015 #15

    haruspex

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    Quite so, but it is a case that needs to be handled.
    Of course, you can probably collapse all the cases into a single formula by judicious use of min and max functions, but I don't think that assists in understanding.
     
  17. Jan 6, 2015 #16

    Ray Vickson

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    Don't forget that for ##\psi \neq 0## the shooter is aiming off center, so the plane containing the target is at a slant. That makes the overlap probabilities a bit trickier to find (although the number of cases is not changed).
    Don't forget that when ##\psi \neq 0## the shooter's "aiming cone" is off-center, so the target's plane is at an angle to the aiming cone's center. That complicates the computation of overlap probabilities a bit, but the number of cases is still the same.
     

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  18. Jan 6, 2015 #17

    haruspex

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    How does that make it harder? It's just a shift in the range of angles, no?
     
  19. Jan 6, 2015 #18

    Ray Vickson

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    Well, yes, but the x-locations of the target ends must be translated into angles, and that involves a small bit of trigonometry. After that, it is the same as before.
     
  20. Jan 6, 2015 #19

    haruspex

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    What x locations? We're given d=Dtanθ, so we can work entirely with θ, Ψ, Φ.
     
  21. Jan 6, 2015 #20

    Ray Vickson

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    OK, I missed that.
     
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