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Probability: having trouble with explanations

  1. Sep 26, 2014 #1
    1. The problem statement, all variables and given/known data
    I GOT all values...but I have trouble explaining.....if someone could suggest how, thank you! Please check the red fonts which shows where I got stuck!

    Suppose we call x the position of the boy and each step he takes (east or west ONLY) is of 1m. So, for an "n" step random walk he takes, the n+1 possible positions that the boy can end up at are given by:

    x = 2k +n, where k = 0, 1 ...n

    and the likelihood of ending up at any one of these is determined by the probability: p(k) = 1/(2^n) (n k) where (n k) = n! / [(n-k)!(k!)] and k = 0, 1 ... n.

    (a) Explain why the formula p(k) should be true.
    (b) Compute <x> after an n-step walk. Why should <x> have this value?
    (c) compute [<x^2>]^0.5 after n step walk. Suppose all n steps were in the same direction, what would be [<x^2>]^0.5?

    2. Relevant equations
    x = 2k +n, where k = 0, 1 ...n
    p(k) = 1/(2^n) (n k) where (n k) = n! / [(n-k)!(k!)] and C is a normalization constant and k = 0, 1 ... n.

    3. The attempt at a solution
    a) B/c Sigma (k=0 to n) p(k) = 1 for normalization, if n =1, supposing x = -1, k would be equal to 1...giving a p(1) = 1/2 (1/1) = 1/2. This is the explanation I put...but, I don't understand in the case when n=1, x = 1, in which k = 0, giving p(0) = 1/2 (1/0) = ????. Could someone hint why the formula is true?

    (b) I obtained the value by:
    <k> = Sigma (k=0 to n) k*p(k)
    = Sigma k (n k) 1/(2^n)
    = 1/(2^n) Sigma k (n k) = 1/(2^n) n * 2^(n-1) = n/2

    therefore...<x> = 2<k> - n
    = 2(n/2) - n = 0....Why did I get the value of zero, if uncertainty is possible????? Is it b/c there is an equal chance of being east or west of the initial position the boy was in the first place?

    (c) Similarly, I derived <k^2> which I got n(n+1)/4. So:
    <x^2> = <(2k+n)^2> = <4k^2 - 4kn + n^2> = 4<k^2> - 4n<k> + n^2 into which I input the results I had from the above...and got <x^2> = n
    So...[<x^2>]^0.5 = n^0.5. However, what should I do if all n steps were in the same direction? Because that is once in all possible probabilities, so, should it be 1/n^0.5? I am really stuck here!
  2. jcsd
  3. Sep 26, 2014 #2

    Ray Vickson

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    The position should be ##x = 2k-n##, not ##2k+n##. Think about it: if he takes ##k## steps to the right and ##n-k## steps to the left, where does he end up?

    You say that in (a) you do not understand the formula for n = 1. Which formula do you mean? Do you want to know why (1 0) = 1? Well, that is just the number of ways of getting 0 heads when you toss a coin one time: you get either H (once) or T (once). In other words, there is just one way to get 1 head and 1 way to get 0 heads. In general, (n 0) = 1 for any integer n because when you toss a coin n times there is just one way of getting 0 heads---namely, by getting all tails.

    BTW: the standard notation for (a b) is either aCb or C(a,b) or ##\binom{a}{b}##. You get the first one by using the "##x_2##" button on the panel at the top of the input page; you get the second one by straight typing; you get the third one using LaTeX.

    In (b) you say you do not understand why ##\langle X \rangle = 0##. Well, what is the MEANING of ##\langle X \rangle ##? It is the average value of X in a large number of identical trials of the situation under study. A mean of 0 just means that going left or right is equally likely. For example, when n = 1 you end up either at x = +1 or x = -1. If you do the experiment 2 million times, then in about 1 million of the experiments you will be at x = 1 and in about 1 million experiments you will be at x = -1. On average, then, your final location will be at zero.

    In (c): the value of ##\langle X^2 \rangle## looks at the average of all the possible ##x^2## values. Only one of the very many outcomes has all steps to the right (or all to the left). The contribution of those two extreme outcomes to ##\langle X^2 \rangle## would be ##2 C(n,0) n^2 / 2^n = 2 n^2/2^n## and so is small compared with the total due to other outcomes.
  4. Sep 26, 2014 #3
    Hello, thank you very much!
    I get (b) and (c)! Also thank you for pointing out how to get the standard notation...I get how to do first two, but, how do I use "Latex"?

    Aside, regarding (a), by formula, I mean I have trouble explaining why p(k) = 1/(2n) (n k) where (n k) = n! / [(n-k)!(k!)] is true--should I verify the function with examples? Or, state that it is b/c of normalization? This is where I am confused.
  5. Sep 27, 2014 #4


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    You may also be under the misapprehension that 0! = 0. By convention, 0!=1. This makes sense because it means that the rule n!*(n+1) = (n+1)! applies even when n=0.
    For p(k), break the problem into two parts. First, how many different sequences of n steps are there? Are they equally likely? What is the probability of each sequence? Next, how many sequences consist of k steps one way and n-k the other?
  6. Sep 27, 2014 #5
    So, 0! = 1. Still, 1! = 1, right?
    Also, about:

    Since the different sequences of n steps there are are 2n, where x + n, x -n both have probability value of 1/2n, whilst the other possibilities are somewhere between two extreme distances, making up of the probability value of (2n -1)/2n.
    As for k = (x + n)/2 after the boy's initial position has moved n steps...the max value k can have is when n steps were in the west direction, as in x - n = 2k -n, in this case, x = 2k....then....because k is dependent on n values, if k = n, then x = 2(n) - n = n....so, how do I get from here? Or, am I getting lost?
  7. Sep 27, 2014 #6

    Ray Vickson

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    I cannot figure out what you are asking, but before you wanted to know why ##P(k) = \binom{n}{k}/2^n##. Is that still what is bothering you? Well, look at some small-scale examples, where you can enumerate all the possibilities. Let the successive outcomes (steps) be labelled as R or L, depending on their direction.

    For n = 3 the possible outcomes are RRR, RRL, RLR, RLL, LLL, LLR, LRL, LRR. There are 8 different outcomes, each having probability 1/8 = 1/2^3. We have P(3R,0L) = P(RRR) = 1/8, P(2R,1L) = P(RRL)+P(RLR) + P(LRR) = 3 * 1/8, P(1R,2L) = P(RLL) + P(LRL) + P(LLR) = 3 * 1/8, and P(0R,3L) = P(LLL) = 1/8. So, the coefficients C(3,k) just count the number of different ways of having k Rs and (3-k) Ls.

    In general, for k Rs and n-k Ls, we have many different "strings", such as RR...RLL...L, RRLR...LLL LR, etc. It is not too difficult to show that the total number of distinct such strings is
    [tex] \text{no. of strings } \; = \binom{n}{k} = \frac{n!}{k! (n-k)!} = \frac{n(n-1) \cdots (n-k+1)}{k!} [/tex]
    In general, if ##p## is the probability of a step to the right and ##q = 1-p## is the probability of a step to the left, the probability of ##k## stepes to the right in ##n## steps altogether is
    [tex] P(k) = \binom{n}{k} p^k q^{n-k}[/tex]
    In the special case ##p = q = 1/2## one gets ##P(k) = \binom{n}{k}/2^n##.
  8. Sep 27, 2014 #7
    Thank you very much.

    One last clarification...
    I think I made a mistake somewhere, for I got <x2> = n, not n2....what I did was:

    <k2> = sigma (k=0 to n) k2 p(k) = (1/2n) sigma k2 (n,k) = 1/(2n) n(n+1)2n-2 = n(n+1)/4
    <k> as done previously = n/2
    <x2> = <(2k-n)2> = 4<k2> - 4<k>n + n2 = n2 + n - 2n2 + n2 = n

    ...so, I got C(n,0) n = 2 (1/2n n!/n!) = 2n/2n, NOT 2n2/2n...

    Could you please point out where I did wrong? If so, merci beaucoup!
  9. Sep 27, 2014 #8

    Ray Vickson

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    ##\langle X^2 \rangle = n## is correct.

    I cannot figure out what your "C(n,0)n" business is all about, or why you should care. The fact is that C(n,0) = 1 so C(n.0)n = n. End of story.
  10. Sep 27, 2014 #9
    Got it!!!!!
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