Figuring out distribution from adding smaller distributions?












1












$begingroup$


Suppose sam wants to know how long it usually takes him to get to work. He wants to know the 50th percentile, 90th percentile, and 99th percentile of how long, in minutes, it takes him to get to work.



Sam's route to work is split up into N segments. For each segment, the time it takes him to traverse that segment is drawn from some distribution over minutes.



Suppose I know the distribution of each segment. That is, for each segment, I know how long it takes to traverse that segment at the 50th percentile, the 90th percentile, the 99th percentile, etc.



How can I figure out the distribution of it takes Sam to get to work from knowing the distributions of the segments?



(Sorry if something doesn't make sense -- edits are welcome. For software engineers: I'm actually trying to figure out how to estimate the latency of a service call composed of several other service calls)






statistics probability-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    If each segment is normal, then you can deduce its mean and variance from the 50th, 90th and 99th percentiles. Then you can add the N means to get the overall mean and (assuming segments are independently distributed) add the N variances to get the overall variance. The sum of normal segments will be normal, From the overall mean and variance you can deduce any desired percentiles. // This is a harder problem if your segments are not normal. // Another approach: if you know the distn's of independent segments it easy to simulate the dist'n of the total, and then find quantiles.
    $endgroup$
    – BruceET
    Jan 18 at 8:50












  • $begingroup$
    In your travelling example Sam may walk quicker if the train in the last section was delayed. Could something similar be true in your actual application? If so, you have correlations and it all gets more complicated.
    $endgroup$
    – user121049
    Jan 18 at 9:04










  • $begingroup$
    Each segment is independent of the others, N is somewhat small, and the segments are not normal, and their distributions are unknown.
    $endgroup$
    – Andre
    Jan 18 at 17:10










  • $begingroup$
    Mean of sum of RVs is sum of individual means. Unfortunately, there is no such relationship for medians. If distributions are symmetrical medians are near means and you might get some sort of rough approx. In my answer, this works roughly for symmetrical normal and uniform dist'ns but not at all for highly skewed exponential distn's. // I'd try to get data to learn something about distn's and then use simulation.
    $endgroup$
    – BruceET
    Jan 18 at 18:57


















1












$begingroup$


Suppose sam wants to know how long it usually takes him to get to work. He wants to know the 50th percentile, 90th percentile, and 99th percentile of how long, in minutes, it takes him to get to work.



Sam's route to work is split up into N segments. For each segment, the time it takes him to traverse that segment is drawn from some distribution over minutes.



Suppose I know the distribution of each segment. That is, for each segment, I know how long it takes to traverse that segment at the 50th percentile, the 90th percentile, the 99th percentile, etc.



How can I figure out the distribution of it takes Sam to get to work from knowing the distributions of the segments?



(Sorry if something doesn't make sense -- edits are welcome. For software engineers: I'm actually trying to figure out how to estimate the latency of a service call composed of several other service calls)






statistics probability-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    If each segment is normal, then you can deduce its mean and variance from the 50th, 90th and 99th percentiles. Then you can add the N means to get the overall mean and (assuming segments are independently distributed) add the N variances to get the overall variance. The sum of normal segments will be normal, From the overall mean and variance you can deduce any desired percentiles. // This is a harder problem if your segments are not normal. // Another approach: if you know the distn's of independent segments it easy to simulate the dist'n of the total, and then find quantiles.
    $endgroup$
    – BruceET
    Jan 18 at 8:50












  • $begingroup$
    In your travelling example Sam may walk quicker if the train in the last section was delayed. Could something similar be true in your actual application? If so, you have correlations and it all gets more complicated.
    $endgroup$
    – user121049
    Jan 18 at 9:04










  • $begingroup$
    Each segment is independent of the others, N is somewhat small, and the segments are not normal, and their distributions are unknown.
    $endgroup$
    – Andre
    Jan 18 at 17:10










  • $begingroup$
    Mean of sum of RVs is sum of individual means. Unfortunately, there is no such relationship for medians. If distributions are symmetrical medians are near means and you might get some sort of rough approx. In my answer, this works roughly for symmetrical normal and uniform dist'ns but not at all for highly skewed exponential distn's. // I'd try to get data to learn something about distn's and then use simulation.
    $endgroup$
    – BruceET
    Jan 18 at 18:57
















1












1








1





$begingroup$


Suppose sam wants to know how long it usually takes him to get to work. He wants to know the 50th percentile, 90th percentile, and 99th percentile of how long, in minutes, it takes him to get to work.



Sam's route to work is split up into N segments. For each segment, the time it takes him to traverse that segment is drawn from some distribution over minutes.



Suppose I know the distribution of each segment. That is, for each segment, I know how long it takes to traverse that segment at the 50th percentile, the 90th percentile, the 99th percentile, etc.



How can I figure out the distribution of it takes Sam to get to work from knowing the distributions of the segments?



(Sorry if something doesn't make sense -- edits are welcome. For software engineers: I'm actually trying to figure out how to estimate the latency of a service call composed of several other service calls)






statistics probability-distributions







share|cite|improve this question









$endgroup$




Suppose sam wants to know how long it usually takes him to get to work. He wants to know the 50th percentile, 90th percentile, and 99th percentile of how long, in minutes, it takes him to get to work.



Sam's route to work is split up into N segments. For each segment, the time it takes him to traverse that segment is drawn from some distribution over minutes.



Suppose I know the distribution of each segment. That is, for each segment, I know how long it takes to traverse that segment at the 50th percentile, the 90th percentile, the 99th percentile, etc.



How can I figure out the distribution of it takes Sam to get to work from knowing the distributions of the segments?



(Sorry if something doesn't make sense -- edits are welcome. For software engineers: I'm actually trying to figure out how to estimate the latency of a service call composed of several other service calls)






statistics probability-distributions




statistics probability-distributions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 18 at 3:41









AndreAndre

61




61












  • $begingroup$
    If each segment is normal, then you can deduce its mean and variance from the 50th, 90th and 99th percentiles. Then you can add the N means to get the overall mean and (assuming segments are independently distributed) add the N variances to get the overall variance. The sum of normal segments will be normal, From the overall mean and variance you can deduce any desired percentiles. // This is a harder problem if your segments are not normal. // Another approach: if you know the distn's of independent segments it easy to simulate the dist'n of the total, and then find quantiles.
    $endgroup$
    – BruceET
    Jan 18 at 8:50












  • $begingroup$
    In your travelling example Sam may walk quicker if the train in the last section was delayed. Could something similar be true in your actual application? If so, you have correlations and it all gets more complicated.
    $endgroup$
    – user121049
    Jan 18 at 9:04










  • $begingroup$
    Each segment is independent of the others, N is somewhat small, and the segments are not normal, and their distributions are unknown.
    $endgroup$
    – Andre
    Jan 18 at 17:10










  • $begingroup$
    Mean of sum of RVs is sum of individual means. Unfortunately, there is no such relationship for medians. If distributions are symmetrical medians are near means and you might get some sort of rough approx. In my answer, this works roughly for symmetrical normal and uniform dist'ns but not at all for highly skewed exponential distn's. // I'd try to get data to learn something about distn's and then use simulation.
    $endgroup$
    – BruceET
    Jan 18 at 18:57




















  • $begingroup$
    If each segment is normal, then you can deduce its mean and variance from the 50th, 90th and 99th percentiles. Then you can add the N means to get the overall mean and (assuming segments are independently distributed) add the N variances to get the overall variance. The sum of normal segments will be normal, From the overall mean and variance you can deduce any desired percentiles. // This is a harder problem if your segments are not normal. // Another approach: if you know the distn's of independent segments it easy to simulate the dist'n of the total, and then find quantiles.
    $endgroup$
    – BruceET
    Jan 18 at 8:50












  • $begingroup$
    In your travelling example Sam may walk quicker if the train in the last section was delayed. Could something similar be true in your actual application? If so, you have correlations and it all gets more complicated.
    $endgroup$
    – user121049
    Jan 18 at 9:04










  • $begingroup$
    Each segment is independent of the others, N is somewhat small, and the segments are not normal, and their distributions are unknown.
    $endgroup$
    – Andre
    Jan 18 at 17:10










  • $begingroup$
    Mean of sum of RVs is sum of individual means. Unfortunately, there is no such relationship for medians. If distributions are symmetrical medians are near means and you might get some sort of rough approx. In my answer, this works roughly for symmetrical normal and uniform dist'ns but not at all for highly skewed exponential distn's. // I'd try to get data to learn something about distn's and then use simulation.
    $endgroup$
    – BruceET
    Jan 18 at 18:57


















$begingroup$
If each segment is normal, then you can deduce its mean and variance from the 50th, 90th and 99th percentiles. Then you can add the N means to get the overall mean and (assuming segments are independently distributed) add the N variances to get the overall variance. The sum of normal segments will be normal, From the overall mean and variance you can deduce any desired percentiles. // This is a harder problem if your segments are not normal. // Another approach: if you know the distn's of independent segments it easy to simulate the dist'n of the total, and then find quantiles.
$endgroup$
– BruceET
Jan 18 at 8:50






$begingroup$
If each segment is normal, then you can deduce its mean and variance from the 50th, 90th and 99th percentiles. Then you can add the N means to get the overall mean and (assuming segments are independently distributed) add the N variances to get the overall variance. The sum of normal segments will be normal, From the overall mean and variance you can deduce any desired percentiles. // This is a harder problem if your segments are not normal. // Another approach: if you know the distn's of independent segments it easy to simulate the dist'n of the total, and then find quantiles.
$endgroup$
– BruceET
Jan 18 at 8:50














$begingroup$
In your travelling example Sam may walk quicker if the train in the last section was delayed. Could something similar be true in your actual application? If so, you have correlations and it all gets more complicated.
$endgroup$
– user121049
Jan 18 at 9:04




$begingroup$
In your travelling example Sam may walk quicker if the train in the last section was delayed. Could something similar be true in your actual application? If so, you have correlations and it all gets more complicated.
$endgroup$
– user121049
Jan 18 at 9:04












$begingroup$
Each segment is independent of the others, N is somewhat small, and the segments are not normal, and their distributions are unknown.
$endgroup$
– Andre
Jan 18 at 17:10




$begingroup$
Each segment is independent of the others, N is somewhat small, and the segments are not normal, and their distributions are unknown.
$endgroup$
– Andre
Jan 18 at 17:10












$begingroup$
Mean of sum of RVs is sum of individual means. Unfortunately, there is no such relationship for medians. If distributions are symmetrical medians are near means and you might get some sort of rough approx. In my answer, this works roughly for symmetrical normal and uniform dist'ns but not at all for highly skewed exponential distn's. // I'd try to get data to learn something about distn's and then use simulation.
$endgroup$
– BruceET
Jan 18 at 18:57






$begingroup$
Mean of sum of RVs is sum of individual means. Unfortunately, there is no such relationship for medians. If distributions are symmetrical medians are near means and you might get some sort of rough approx. In my answer, this works roughly for symmetrical normal and uniform dist'ns but not at all for highly skewed exponential distn's. // I'd try to get data to learn something about distn's and then use simulation.
$endgroup$
– BruceET
Jan 18 at 18:57












1 Answer
1






active

oldest

votes


















0












$begingroup$

Comment continued: Here are three specific examples to illustrate
see that finding the exact quantiles of sums of distributions is not trivial.



Ten normal segments. Suppose each of ten independent segments is
distributed $mathsf{Norm}(mu = 10, sigma=1).$ Then quantiles .5, .9 and .99 for each segment are about 10, 11.3, and 12.3, respectively. (Computations in R.)



qnorm(c(.5,.9,.99), 10, 1)

10.00000 11.28155 12.32635


The sum of ten such segments is distributed $mathsf{Norm}(mu=100, sigma=sqrt{10}).$
The corresponding quantiles of this distribution are about 100, 104, and 107.$



qnorm(c(.5,.9,.99), 100, sqrt(10))

100.0000 104.0526 107.3566


Ten exponential segments. Suppose each of ten independent segments is
distributed $mathsf{Exp}(text{rate} = .1).$ Then quantiles .5, .9 and .99 for each segment are about 6.9, 23, and 46, respectively. (The mean and standard deviation are both $10.)$



qexp(c(.5,.9,.99), .1)

 6.931472 23.025851 46.051702


The sum of ten such segments is distributed $mathsf{Gamma}(text{shape}=10, text{rate}=.1.)$ (The mean is 100 and the variance is 1000.) The corresponding quantiles of the sum are about 97, 142, and 188.



qgamma(c(.5,.9,.99), 10, .1)

96.68715 142.05990 187.83117


Similar approximate results from a simulation:



set.seed(118)

x = replicate(10^6, sum(rexp(10,.1)))  # vector or a million sums of ten

mean(x); var(x); quantile(x, c(.5,.9,.99))

99.97961

1000.581

      50%       90%       99% 

 96.67126 142.06101 188.03925


Sum of a dozen uniform segments.
Suppose each of ten independent segments is
distributed $mathsf{Unif}(0,1).$ Then quantiles .5, .9 and .99 for each segment are .5, .9, and .99, respectively. (The mean is 1/2 and the variance is 1/12.)



qunif(c(.5,.9,.99))

0.50 0.90 0.99


According to the Central Limit Theorem, the sum of 12 such segments is distributed nearly as $mathsf{Norm}(mu=6, sigma=1)$ From simulation, the approximate corresponding quantiles of the sum are about 6, 7.3, and 8.3.



set.seed(2019)

x = replicate(10^6, sum(runif(12)))

mean(x); var(x); quantile(x, c(.5,.9,.99))

6.001354

1.000741

     50%      90%      99% 

6.002158 7.289854 8.310085


Note: In general, if the segments are independent and identically distribututed with known mean and variance, and there are enough of them that the Central Limit
Theorem applies, then you might find the mean and variance of the nearly normal sum, and from them the desired quantiles.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077809%2ffiguring-out-distribution-from-adding-smaller-distributions%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Comment continued: Here are three specific examples to illustrate
    see that finding the exact quantiles of sums of distributions is not trivial.



    Ten normal segments. Suppose each of ten independent segments is
    distributed $mathsf{Norm}(mu = 10, sigma=1).$ Then quantiles .5, .9 and .99 for each segment are about 10, 11.3, and 12.3, respectively. (Computations in R.)



    qnorm(c(.5,.9,.99), 10, 1)
    
10.00000 11.28155 12.32635


    The sum of ten such segments is distributed $mathsf{Norm}(mu=100, sigma=sqrt{10}).$
    The corresponding quantiles of this distribution are about 100, 104, and 107.$



    qnorm(c(.5,.9,.99), 100, sqrt(10))
    
100.0000 104.0526 107.3566


    Ten exponential segments. Suppose each of ten independent segments is
    distributed $mathsf{Exp}(text{rate} = .1).$ Then quantiles .5, .9 and .99 for each segment are about 6.9, 23, and 46, respectively. (The mean and standard deviation are both $10.)$



    qexp(c(.5,.9,.99), .1)
    
 6.931472 23.025851 46.051702


    The sum of ten such segments is distributed $mathsf{Gamma}(text{shape}=10, text{rate}=.1.)$ (The mean is 100 and the variance is 1000.) The corresponding quantiles of the sum are about 97, 142, and 188.



    qgamma(c(.5,.9,.99), 10, .1)
    
96.68715 142.05990 187.83117


    Similar approximate results from a simulation:



    set.seed(118)
    
x = replicate(10^6, sum(rexp(10,.1)))  # vector or a million sums of ten
    
mean(x); var(x); quantile(x, c(.5,.9,.99))
    
99.97961
    
1000.581
    
      50%       90%       99% 
    
 96.67126 142.06101 188.03925


    Sum of a dozen uniform segments.
    Suppose each of ten independent segments is
    distributed $mathsf{Unif}(0,1).$ Then quantiles .5, .9 and .99 for each segment are .5, .9, and .99, respectively. (The mean is 1/2 and the variance is 1/12.)



    qunif(c(.5,.9,.99))
    
0.50 0.90 0.99


    According to the Central Limit Theorem, the sum of 12 such segments is distributed nearly as $mathsf{Norm}(mu=6, sigma=1)$ From simulation, the approximate corresponding quantiles of the sum are about 6, 7.3, and 8.3.



    set.seed(2019)
    
x = replicate(10^6, sum(runif(12)))
    
mean(x); var(x); quantile(x, c(.5,.9,.99))
    
6.001354
    
1.000741
    
     50%      90%      99% 
    
6.002158 7.289854 8.310085


    Note: In general, if the segments are independent and identically distribututed with known mean and variance, and there are enough of them that the Central Limit
    Theorem applies, then you might find the mean and variance of the nearly normal sum, and from them the desired quantiles.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Comment continued: Here are three specific examples to illustrate
      see that finding the exact quantiles of sums of distributions is not trivial.



      Ten normal segments. Suppose each of ten independent segments is
      distributed $mathsf{Norm}(mu = 10, sigma=1).$ Then quantiles .5, .9 and .99 for each segment are about 10, 11.3, and 12.3, respectively. (Computations in R.)



      qnorm(c(.5,.9,.99), 10, 1)
      
10.00000 11.28155 12.32635


      The sum of ten such segments is distributed $mathsf{Norm}(mu=100, sigma=sqrt{10}).$
      The corresponding quantiles of this distribution are about 100, 104, and 107.$



      qnorm(c(.5,.9,.99), 100, sqrt(10))
      
100.0000 104.0526 107.3566


      Ten exponential segments. Suppose each of ten independent segments is
      distributed $mathsf{Exp}(text{rate} = .1).$ Then quantiles .5, .9 and .99 for each segment are about 6.9, 23, and 46, respectively. (The mean and standard deviation are both $10.)$



      qexp(c(.5,.9,.99), .1)
      
 6.931472 23.025851 46.051702


      The sum of ten such segments is distributed $mathsf{Gamma}(text{shape}=10, text{rate}=.1.)$ (The mean is 100 and the variance is 1000.) The corresponding quantiles of the sum are about 97, 142, and 188.



      qgamma(c(.5,.9,.99), 10, .1)
      
96.68715 142.05990 187.83117


      Similar approximate results from a simulation:



      set.seed(118)
      
x = replicate(10^6, sum(rexp(10,.1)))  # vector or a million sums of ten
      
mean(x); var(x); quantile(x, c(.5,.9,.99))
      
99.97961
      
1000.581
      
      50%       90%       99% 
      
 96.67126 142.06101 188.03925


      Sum of a dozen uniform segments.
      Suppose each of ten independent segments is
      distributed $mathsf{Unif}(0,1).$ Then quantiles .5, .9 and .99 for each segment are .5, .9, and .99, respectively. (The mean is 1/2 and the variance is 1/12.)



      qunif(c(.5,.9,.99))
      
0.50 0.90 0.99


      According to the Central Limit Theorem, the sum of 12 such segments is distributed nearly as $mathsf{Norm}(mu=6, sigma=1)$ From simulation, the approximate corresponding quantiles of the sum are about 6, 7.3, and 8.3.



      set.seed(2019)
      
x = replicate(10^6, sum(runif(12)))
      
mean(x); var(x); quantile(x, c(.5,.9,.99))
      
6.001354
      
1.000741
      
     50%      90%      99% 
      
6.002158 7.289854 8.310085


      Note: In general, if the segments are independent and identically distribututed with known mean and variance, and there are enough of them that the Central Limit
      Theorem applies, then you might find the mean and variance of the nearly normal sum, and from them the desired quantiles.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Comment continued: Here are three specific examples to illustrate
        see that finding the exact quantiles of sums of distributions is not trivial.



        Ten normal segments. Suppose each of ten independent segments is
        distributed $mathsf{Norm}(mu = 10, sigma=1).$ Then quantiles .5, .9 and .99 for each segment are about 10, 11.3, and 12.3, respectively. (Computations in R.)



        qnorm(c(.5,.9,.99), 10, 1)
        
10.00000 11.28155 12.32635


        The sum of ten such segments is distributed $mathsf{Norm}(mu=100, sigma=sqrt{10}).$
        The corresponding quantiles of this distribution are about 100, 104, and 107.$



        qnorm(c(.5,.9,.99), 100, sqrt(10))
        
100.0000 104.0526 107.3566


        Ten exponential segments. Suppose each of ten independent segments is
        distributed $mathsf{Exp}(text{rate} = .1).$ Then quantiles .5, .9 and .99 for each segment are about 6.9, 23, and 46, respectively. (The mean and standard deviation are both $10.)$



        qexp(c(.5,.9,.99), .1)
        
 6.931472 23.025851 46.051702


        The sum of ten such segments is distributed $mathsf{Gamma}(text{shape}=10, text{rate}=.1.)$ (The mean is 100 and the variance is 1000.) The corresponding quantiles of the sum are about 97, 142, and 188.



        qgamma(c(.5,.9,.99), 10, .1)
        
96.68715 142.05990 187.83117


        Similar approximate results from a simulation:



        set.seed(118)
        
x = replicate(10^6, sum(rexp(10,.1)))  # vector or a million sums of ten
        
mean(x); var(x); quantile(x, c(.5,.9,.99))
        
99.97961
        
1000.581
        
      50%       90%       99% 
        
 96.67126 142.06101 188.03925


        Sum of a dozen uniform segments.
        Suppose each of ten independent segments is
        distributed $mathsf{Unif}(0,1).$ Then quantiles .5, .9 and .99 for each segment are .5, .9, and .99, respectively. (The mean is 1/2 and the variance is 1/12.)



        qunif(c(.5,.9,.99))
        
0.50 0.90 0.99


        According to the Central Limit Theorem, the sum of 12 such segments is distributed nearly as $mathsf{Norm}(mu=6, sigma=1)$ From simulation, the approximate corresponding quantiles of the sum are about 6, 7.3, and 8.3.



        set.seed(2019)
        
x = replicate(10^6, sum(runif(12)))
        
mean(x); var(x); quantile(x, c(.5,.9,.99))
        
6.001354
        
1.000741
        
     50%      90%      99% 
        
6.002158 7.289854 8.310085


        Note: In general, if the segments are independent and identically distribututed with known mean and variance, and there are enough of them that the Central Limit
        Theorem applies, then you might find the mean and variance of the nearly normal sum, and from them the desired quantiles.






        share|cite|improve this answer











        $endgroup$



        Comment continued: Here are three specific examples to illustrate
        see that finding the exact quantiles of sums of distributions is not trivial.



        Ten normal segments. Suppose each of ten independent segments is
        distributed $mathsf{Norm}(mu = 10, sigma=1).$ Then quantiles .5, .9 and .99 for each segment are about 10, 11.3, and 12.3, respectively. (Computations in R.)



        qnorm(c(.5,.9,.99), 10, 1)
        
10.00000 11.28155 12.32635


        The sum of ten such segments is distributed $mathsf{Norm}(mu=100, sigma=sqrt{10}).$
        The corresponding quantiles of this distribution are about 100, 104, and 107.$



        qnorm(c(.5,.9,.99), 100, sqrt(10))
        
100.0000 104.0526 107.3566


        Ten exponential segments. Suppose each of ten independent segments is
        distributed $mathsf{Exp}(text{rate} = .1).$ Then quantiles .5, .9 and .99 for each segment are about 6.9, 23, and 46, respectively. (The mean and standard deviation are both $10.)$



        qexp(c(.5,.9,.99), .1)
        
 6.931472 23.025851 46.051702


        The sum of ten such segments is distributed $mathsf{Gamma}(text{shape}=10, text{rate}=.1.)$ (The mean is 100 and the variance is 1000.) The corresponding quantiles of the sum are about 97, 142, and 188.



        qgamma(c(.5,.9,.99), 10, .1)
        
96.68715 142.05990 187.83117


        Similar approximate results from a simulation:



        set.seed(118)
        
x = replicate(10^6, sum(rexp(10,.1)))  # vector or a million sums of ten
        
mean(x); var(x); quantile(x, c(.5,.9,.99))
        
99.97961
        
1000.581
        
      50%       90%       99% 
        
 96.67126 142.06101 188.03925


        Sum of a dozen uniform segments.
        Suppose each of ten independent segments is
        distributed $mathsf{Unif}(0,1).$ Then quantiles .5, .9 and .99 for each segment are .5, .9, and .99, respectively. (The mean is 1/2 and the variance is 1/12.)



        qunif(c(.5,.9,.99))
        
0.50 0.90 0.99


        According to the Central Limit Theorem, the sum of 12 such segments is distributed nearly as $mathsf{Norm}(mu=6, sigma=1)$ From simulation, the approximate corresponding quantiles of the sum are about 6, 7.3, and 8.3.



        set.seed(2019)
        
x = replicate(10^6, sum(runif(12)))
        
mean(x); var(x); quantile(x, c(.5,.9,.99))
        
6.001354
        
1.000741
        
     50%      90%      99% 
        
6.002158 7.289854 8.310085


        Note: In general, if the segments are independent and identically distribututed with known mean and variance, and there are enough of them that the Central Limit
        Theorem applies, then you might find the mean and variance of the nearly normal sum, and from them the desired quantiles.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 18 at 10:29

























        answered Jan 18 at 9:51









        BruceETBruceET

        35.6k71440




        35.6k71440






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077809%2ffiguring-out-distribution-from-adding-smaller-distributions%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
            Read more

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
            Read more

            Dobbiaco

            Image
            Dobbiaco comune ( IT ) Dobbiaco ( DE ) Toblach Dobbiaco Vecchia Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Guido Bocher (lista civica Indipendenti-Unabhängige) dal 10-5-2015 (2º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Coordinate: 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Altitudine 1 213 m s.l.m. Superficie 125,42 km² Abitanti 3 356 [2] (31-7-2017) Densità 26,76 ab./km² Frazioni Carbonin ( Schluderbach ), Santa Maria ( Aufkirchen ), Valle San Silvestro ( Wahlen ) Comuni confinanti Auronzo di Cadore (BL), Braies, Cortina d'Ampezzo (BL), Innervillgraten (AT-7), San Candido, Sesto, Valle di Casies, Villabassa Altre informazioni Cod. postale 39034 Prefisso 0474 Fuso...
            Read more