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Most probable value of a random variable?

  1. Oct 2, 2012 #1
    A problem in this book asks for the most probable value of a random variable [itex]x[/itex]. As far as I know, if a random variable has "most probable value" then it isn't a random variable.

    The problem is attached. It is the second question in part b.

    Could the answer be that there is no most probable value? Or do I not understand exactly what a random variable is.

    If it's asking at what value of [itex]x[/itex] is [itex]p(x)[/itex] greatest, that would of course be at [itex]x = 0[/itex] but I don't think that is what the question is asking.
     

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    Last edited: Oct 2, 2012
  2. jcsd
  3. Oct 2, 2012 #2
    Thought about it more.

    Could it be that it is asking for [itex]p(\bar{x})[/itex] since it asked me to find the mean value of [itex]x[/itex]?
     
  4. Oct 2, 2012 #3

    mathman

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    The most probable value is the x where p(x) is maximum.
     
  5. Oct 2, 2012 #4
    The "most probable value" should not be confused with the "expectation": [itex]E(x) = \bar {x} [/itex]. The uniform distribution has an expectation but no "most probable value" since all values over a discrete range are equally probable. Also, in many distributions the expectation is not the most probable value when such a value exists. In a unimodal distibution, the most probable value is the mode. For the normal distribution, the mean, median and mode are located at the same point.

    EDIT: I'm speaking in terms of discrete distributions where n approaches infinity, but we can still talk about mapping from a point x to a probability space. In a true continuum, the probability of any point x is zero. For symmetric continuous probability distributions, we can usually use the half distribution probability density.
     
    Last edited: Oct 2, 2012
  6. Oct 3, 2012 #5
    That's true. But then we can use the maximum of the probability density function, which is a fancy way of saying we draw the graph of the probability and find the highest point on that. So it is intuitively very similar.
     
  7. Oct 3, 2012 #6
    Well, strictly speaking the maximum probability density is always 1. You have to find the upper limit of integration (the maximum point) analytically afaik. For multimodal functions, it becomes more difficult. Of course you can always plot the function to obtain an approximate result by inspection or use the plot to find the upper limit of integration (the global maximum point).
     
    Last edited: Oct 3, 2012
  8. Oct 6, 2012 #7

    haruspex

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    No, it can be anything > 0, even greater than 1.
    For a continuous distribution, it doesn't really make sense to ask for the most probable value, for the reasons already discussed. The max of p(x) is the most reasonable interpretation. Denjay, why do you think that makes it not a random variable? A thing is random if it is not deterministic.
     
    Last edited: Oct 6, 2012
  9. Oct 6, 2012 #8
    The maximum probability density on a PDF is one for the purposes of the OP. The uniform distribution has no mode.

    It may make sense to find the mode or the maximum mode of the PDF.
     
    Last edited: Oct 6, 2012
  10. Oct 6, 2012 #9

    chiro

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    In statistics we have a thing called Highest Posterior Density and in the context of what SW VandeCarr said, this makes sense for a non-zero interval where you have an actual probability and not just a point estimate for MLE estimation (or a similar technique to get maximum values).
     
  11. Oct 7, 2012 #10

    haruspex

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    I said nothing about uniform distributions. In the context of the OP, the max density is A, which can be > 1.
     
  12. Oct 7, 2012 #11
    Please give an example in terms of actual probabilities..
     
    Last edited: Oct 7, 2012
  13. Oct 7, 2012 #12

    haruspex

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    2e-2x, x>=0; 0 for x<0.
     
  14. Oct 7, 2012 #13
    That's not an actual probability. A probability cannot be greater than one. So called probability densities greater than one are not actual probabilities and have no relevance to the OP's question.
     
  15. Oct 7, 2012 #14

    haruspex

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    The probability density is not a probability. It is the value of the PDF at that x value. Since the agreed(?) interpretation of the OP question is in terms of probability density, not actual probabilities, it is relevant. If it is not to be interpreted as max prob density then the OP question is meaningless for a continuous distribution.
     
  16. Oct 7, 2012 #15
    Agreed. The OP asked for the maximum value of p(x) which means the probability of x.
     
    Last edited: Oct 7, 2012
  17. Oct 7, 2012 #16

    haruspex

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    No, the OP asked for the most probable value, and it defined p(x) to be the pdf.
     
  18. Oct 7, 2012 #17
    It's a very reasonable assumption that the OP was thinking in terms of the standard distributions that are used in statistical applications. Integrating the PDFs of these distributions from negative to positive infinity must yield 1 in order to satisfy the axioms of probability theory. You know this. The kinds of distributions you refer to would not appear to have any applications in the kinds of statistical/probability problems that are generally discussed in this forum.

    Given we are referring to these kinds of standard distributions, the probability density is a useful tool when dealing with PDFs. Even when dealing with discrete distributions with large n, using the continuous approximation is often considered preferable.

    Finding the mode of a unimodal standard PDF is a straightforward analytic procedure. You know this too. I don't think the line you're pursuing is helpful to the OP. If you persist, I will refer this thread to the moderators.
     
    Last edited: Oct 7, 2012
  19. Oct 7, 2012 #18

    Mute

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    No, you are wrong and haruspex is correct. The attachment in the OP explicitly states that ##p(x)## is a probability density. It is NOT a probability.

    So, this should clear up the miscommunication, unless you also happen to mistakenly believe that a probability density cannot have a maximum value greater than one. I cannot tell from your post if you believe that or if there is just a misunderstanding of what ##p(x)## meant, but in case it is the former: A valid probability density function can absolutely have a maximum value greater than 1. The total area under the curve must be 1, but that does not at all imply the function itself cannot be greater than 1.

    Consider a uniform distribution of random variables that can take value only within a very tiny range, say 0.9 to 1.1. The probability density function is p(x) = 5. The area under this curve is 1:

    $$\int_{0.9}^{1.1} dx~p(x) = 5 \int_{0.9}^{1.1} dx = 5 (1.1-0.9) = 5 (0.2) = 1.$$

    This is a perfectly valid, normalized probability density function. Because the width is so small, the height has to be large in order for the area to come out to 1.

    Similarly, take any Gaussian and give it a very small standard deviation. You will find the probability density function exceeds 1 when x is equal to the mean, but the area under the curve will never exceed 1. Again, this is because the peak height has to compensate for the fact that there is very little area under the tails of the Gaussian curve when the standard deviation is very small.

    To the OP:

    I would interpret the question as expecting you to give the answer "x=0" as the "most probable value" or the "most likely value". Although strictly speaking the probability of getting the value is 0 for any given point, if you draw from this distribution many, many times and generate a histogram which approximates the (unnormalized) distribution, then you will see that you drew many more points in your "x~0" bin than any other points. If you drew from the distribution infinitely many times so that you could make your bins infinitely thin, the greatest fraction of the points that you drew will correspond to "x=0". So, in this sense, the value "x=0" is the most probable value.

    Another way to see it is to realize that if you draw from the probability density ##p(x)## infinitely many times, ##f(x_1,x_2) = (Ae^{-\lambda x_1})/(Ae^{-\lambda x_2}) = e^{-\lambda(x_1-x_2)}## is the fraction of times you drew ##x_1## compared to ##x_2##. For ##x_1 = 0 \neq x_2##, you can see that this fraction is always greater than 1, so there are always more draws of "x=0" than any other point. (You don't have to make this argument for every problem like this, it's just to illustrate the sense in which the value of x that maximizes p(x) is the "most probable value").
     
    Last edited: Oct 8, 2012
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