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Probability of sum of 5 independant random variables

  1. Jan 10, 2012 #1
    Hi. I would like to find out the probability distributions function of the sum of 5 independant random variables. They are a sum of errors: 1%, 1%, 0.1%, 0.1%, 1%.
    I think this is the convolution of all these.
    So the limits are +/- 3.2%

    I know the convolution of 2 square pulses becomes a triangle, but i'm unsure about lots of them.

    I was also reading the central limit theorem that says convolution of many random variables aproaches a normal distribution, but I don't know what the height or width of it would be.
  2. jcsd
  3. Jan 10, 2012 #2

    Simon Bridge

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    since they are independant, you sum the variences. Which means you need the measurements as well.
  4. Jan 10, 2012 #3
    I'm not sure I understand. Maybe I am asking wrong.

    I have a start value. say 100%, then I have an error of +/-1%
    Then I have an equal probability of getting from 99 to 101%

    If I independantly add another error of 1%, the distribution is like a triangle from 98% to 102%. So the probability of being within 1% is maybe 75% at a guess.

    The distribution of each of these is even, so I don't think the varience is necessary. It is not 1% of some number.

    Maybe I should say
    -1 to 1, -1 to 1, -0.1 to 0.1, -0.1 to 0.1, -1 to 1
    don't wory about the % sign.
  5. Jan 10, 2012 #4

    Simon Bridge

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    If you make a measurement x, the statistical uncertainty on that measurement is the standard-deviation of the distribution of possible measurements X. We treat the actual x as if it were the mean of a normal distribution.

    So we write: X is x ± σx units.

    We can also express the error as a percentage - px= 100σx/x

    When you add two measurements, z=x+y, then z is the new mean, and the new standard deviation is given by the sum of the squares of the undividual means thus:

    σz2= σx2+ σy2

    so, unless all your measurements have the same value, you need them to find the variences of each distribution.
    You seem to be asking to find the percentage error on the sum knowing only the percentage errors of the terms, and not the terms themselves.

    The only way the % sign does not matter is if you are talking about the percentile error on a computer percentage - like in political polls, where party A gets 34% and the error is 2% - meaning "2 percentage points" - so the actual support is supposed to be 95% likely to lie between 32% and 38%.

    Perhaps if you showed me the actual problem?
    Last edited: Jan 10, 2012
  6. Jan 11, 2012 #5
    Thanks for the explanation. I did learn some probability but I wasn't good at it at the time.

    I have a electronic circuit that amplifies AC signal. the first stage is an opamp with 1% resistors. the gain is -rf/rs. The worst case is that the rf resistor is 101% of it's value and the rs resistor is 99% of it's value then 1.01/ 0.99 is approx 1.02. I don't know what the distribution of the probability of the components is. I was going to assume an even distribution.

    The next stage is with 0.1% resistors, and a 1% on the load.

    I wanted to get an estimate of the likelihood of the final amplifier being out by 3.2% or if it is reasonable to assume that it will be within 2%.

    I thought knowing the probability distribution function would help. I'm sure this is a common problem in engineering I wanted to know how to solve it so to know how to solve future similar problems.

    From what you are saying, the standard deviation i think is sqrt((1-(-1))/12) = sqrt(1/3)
    Therefore to add the squares: = 1/3 + 1/3 = 2/3 therefore standard deviation is sqrt(2/3).
    Then I would do the same for the other values. is this correct?
  7. Jan 11, 2012 #6

    Simon Bridge

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    Resisters have special statistics because the tolerance is not the same as the standard error[1]
    ... all resisters are manufactured so their values are distributed approximately R ~ (r,r/10) ... but they have been tested and divides into groups, so the
    0.1% tolerance resisters are those within 0.1% of the mean;
    1% tolerance resisters are those which were between 0.1% and 1% of the mean.

    Both the 1% and 0.1% can be approximated as flat - which changes the stats, but I'll use the standard normal anyway to avoid awkward math.

    Example: a 100k resister at 1% tolerance would vary between 101k and 99k
    A 50k resister at 1% tolerance would vary between 50k5 and 49k5.
    Put them in series means the resulting resistance varies between 148k5 and 152k5 but the distribution will be approximately R ~ N(150k,5k) within those limits.[2]

    But you are not adding resistors - you are dividing the: the gain is [itex]g=-r_f/r_s[/itex]
    This is a ratio of independent uncertain values, not a sum or difference.
    The rule is different for multiplication and division ... in this case you do the sum-of-squares thing on the percentage errors :)

    [itex]p_g = \sqrt{p_f^2 + p_s^2}[/itex]
    ... the distribution would then vary by [itex]10\sqrt{2}%[/itex] of whatever the actual gain is.
    ... the tolerance range will be 2% (sum of tolerances) from the mean.

    (recall there's a hole at the 0.1% limit though).

    To summarize: manufactured resistors are special...
    ... for the gain found from the ratio of two 1% resistances:
    effective tolerance of the gain is 2%
    distribution of values within the tolerance range has standard deviation 14% of the nominal value of the gain. "Nominal value" because there is likely zero chance of finding the ratio of 1% resistors within 0.2% of this value.
    ... for the ratio of two 0.1% resistances:
    that's 0.2% and 14%.

    [1] I actually tested this ... different manufacturers may do this differently, but I remember buying a lot of 10% resisters and painstakingly going through them with an ohmmeter to sort of the 1% ones, only to find there weren't any :( [3]
    [2] the "~" (tilde) in this context means "distributed according to", and N(m,s) is the normal distribution with a mean of m and a standard deviation of s.
    [3] there is also an uncertainty in the tolerance cut-off values - but this is very much smaller than the 10% for the manufactured distribution of values. You should try this with a sample of stock resisters.
  8. Jan 11, 2012 #7
    OK thanks!

    so I can say that my standard deviation is:
    sqrt(0.01^2 + 0.01^2 + 0.001^2 + 0.001^2 + 0.01^2)
    = sqrt(0.00302)
    =0.0174 = 1.74%

    So I now wonder what percentage of the system would be within 1.74%? could I say that 68% of results are within 1.74%. Can I figure out what 90% of the results are within?

    ps. I think you meant 1.4% and not 14%.
  9. Jan 11, 2012 #8

    Simon Bridge

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    Nah - remember, the overall distribution has a 10% standard deviation regardless of the tolerance. Your resisters have already been sorted from that distribution to get the tolerances.

    You have to normalize the distribution within the range - easier to do if you assume it's flat within the tolerances.

    So a 100k resister with 1% tolerance is approximately equally likely to be anywhere in
    99k<R<99.9 and 100.1<R<101 (Don't confuse tolerance with error.)

    The same resister with a 0.1% tolerance is approximately equally likely between 99.9 and 100.1, with no gap in the middle if that is the highest tolerance resister available.

    The gains will inherit tolerances from the resister ratios - 2% and 0.2% respectively.
    Again approximating to flat distributions a 2% tolerance gain is equally likely to have values between between 98% and 102% of the nominal gain except at the hole (what happens at the hole?)

    If you have a lot of these though, the mean value theorem means the final distributions will be more normal.
  10. Jan 11, 2012 #9


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    Do you mean the Central Limit Theorem?
  11. Jan 12, 2012 #10

    Simon Bridge

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    Good grief - you know, I'd been making small mistakes in names like that all night?!
    Thank you, well spotted.
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