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> Many samples needed, especially in high dimensions This isn't true. For Monte Carlo sampling, the convergence of unbiased estimators (for example the expect

> Many samples needed, especially in high dimensions This isn't true. For Monte Carlo sampling, the convergence of unbiased estimators (for example the expectation) is independent of the dimension of the state space. In fact, this is exactly the reason to prefer Monte Carlo integration over, say, a Riemann sum.

glial 10y ago
This is true, but in my experience sampling can take longer with higher-dimensional models, even with the same number of samples. So from an efficiency point of view, VI is still great compared to sampling.
murbard2 10y ago
That's true in the asymptotic, but the dimension find a way to rear its head in the constant to the big O through the variance of your estimator. It's still better than Riemann integration, but it sucks nonetheless. For instance, integrate the function f(x1,x2,...xd) = 6^d * x1 * (1-x1) * ... * xd * (1-xd) on the d-dimensional unit cub (the answer is 1, for all d). the expected variance for a single point estimator is -1+(6/5)^d, which increases exponentially in the number of dimensions. That's the multiplicative constant to your big O. For a probabilistic example let x_i ~ N(0,1) for i in 1..d and let s = Sum x_i. Try estimating the probability that -0.1 < s < 0.1 by Monte-Carlo sampling. If most of the probability distribution mass is located near a manifold of lower dimensions, as tends to be the case for natural data, your variance will be huge. MCMC, HMC both improve this state of affair by letting you walk or "glide" on the manifold, but you still have to contend with curvature, with multiple modes etc.
- yummyfajitas 10y ago
  How does MCMC improve this state of affairs? From what I can see, MCMC only worsens things since you can't be sure you are sampling the entire manifold.
  - murbard2 10y ago
    That's a good question. If your distribution isn't multimodal but still suffers from having its mass concentrated in one place, and if the variable you're integrating isn't fat-tailed, then MCMC will help you by sticking around the right area once it finds it.
    - yummyfajitas 10y ago
      I don't even think that's true. Consider a distribution in R^2 that is concentrated on [0,1] x [0,100]. If you have an O(1) jump size it'll take time O(100^2) for the MCMC to fully sample the support of this distribution. If you had a jump size of O(100), then you'll be rejecting 99% of your jumps due to the narrowness in the x-direction. This is hardly an uncommon scenario.
      - murbard2 10y ago
        It's easy to design examples that trip MCMC, but they also trip straight Monte Carlo integration even more. You can, of course, construct examples where MCMC will do worse than simple Monte Carlo integration, but these are uncommon. They mostly illustrate the difficulty of picking an appropriate jump size.
        yummyfajitas 10y ago
        If you directly draw samples from the underlying distribution, you don't have to worry at all about whether the markov chain or other process (e.g. HMC) is fully sampling the distribution. MC has no rejections and always samples the entire distribution. You simply don't need to worry about trajectories not going everywhere they should.
      - deleted 10y ago
        [deleted]
- zump 10y ago
  What course did you study to get this knowledge?
  - AlexCoventry 10y ago
    Any course on the mathematics of Bayesian inference will introduce you to the necessary tools.
  - murbard2 10y ago
    AlexCoventry is right, these are "the mathematics of Bayesian inference", to unpack a bit my recommendation is: - Get a strong grip on linear algebra and euclidean spaces. You should be able to have an intuitive feel for the key theorems (e.g. Cayley-Hamilton, the Spectral theorem) and should be comfortable proving them (at least once). The point isn't that you need to check that they're correct (they are), but if you can prove them, it means you've learned a certain amount of prerequisite knowledge and gained enough familiarity with the topic. They aren't just theorems you apply, they make sense and you understand the intuition behind them. - Get the same feel for multivariate calculus. The intuition there is generally easier to acquire than for linear algebra, but you need to be comfortable with the mechanics of it. Learn to prove your results rigorously, but also to quickly derive formulas by treating infinitesimals as variables, like a physicist. - Study integration, measures, distributions and the fundamentals of probability. If the course talks about sigma-algebras, you're in the right place. Finally study Baseyian statistics, Monte-Carlo Integration, Markov-Chain Monte-Carlo, some information theory You don't need that level of rigor and that level of fundamentals to do machine learning. Some linear algebra, some calculus and some probability theory that you pick along the way will generally do. However, I think if you're interested in ML it is worth the effort because it will make most of the math seamless. This is a lot of math to learn, but it's not particularly "advanced" math. The underlying intuitions are relatively concrete and a lot of the procedures are relatively mechanical.
    - zump 10y ago
      Thanks for the wonderful reply. How long do you estimate it took for you to get to the level you are now?
srean 10y ago
murbard2 and glial have pointed out some caveats. Another thing that ties down Monte Carlo sampling at times is when it is not easy to generate the samples required for Monte Carlo. A common route is to use a Markov chain sampler, hence MCMC, but that's far from a solved problem. Its very hard to reason and test that the Markov chain mixes fast, and if it does not it would take exponential time to generate the samples from its stationary distribution. Coming up with schemes better than a naïve Markov chain is an intellectual industry in itself.
- murbard2 10y ago
  HMC has better asymptotics than MCMC. If your probability surface is differentiable you should always use that. Riemannian MC takes curvature into account but you need to start storing Hessians which isn't practical. Two missing pieces would be: - An SGD equivalent for sampling "tall" data. The generalized Poisson sampler can do that, but the variance is crappy. - An L-BFGS like version of Riemannian MC, to get rid of the Hessians. Parallel tempering should also be exploited more. That said, at the end of the day, the ideal sampler would be able to reason about the distribution as program and not just as a mathematical black box. It should build tractable approximations intelligently and use those to bootstrap exact sampling schemes. I think we're going to see a wealth of better samplers come out in the next decade, following the path of combinatorial optimization towards preserving the structure of the programs.
  - marmaduke 10y ago
    > you need to start storing Hessians aren't there methods for online sparse estimates of the Hessian? I'd expect a lot of "large" problems for which RMC is useful would have sparse structures. > ideal sampler would be able to reason about the distribution as program Are you a developer of Stan? If not, you might be interested. http://mc-stan.org http://mc-stan.org _edit_ by online estimate of a Hessian, I meant online numerical approximation based on the sequence of Jacobians.
    - jmalicki 10y ago
      "An L-BFGS like version of Riemannian MC, to get rid of the Hessians." - that's what he said

glial 10y ago

This is true, but in my experience sampling can take longer with higher-dimensional models, even with the same number of samples. So from an efficiency point of view, VI is still great compared to sampling.

murbard2 10y ago

That's true in the asymptotic, but the dimension find a way to rear its head in the constant to the big O through the variance of your estimator. It's still better than Riemann integration, but it sucks nonetheless. For instance, integrate the function f(x1,x2,...xd) = 6^d * x1 * (1-x1) * ... * xd * (1-xd) on the d-dimensional unit cub (the answer is 1, for all d). the expected variance for a single point estimator is -1+(6/5)^d, which increases exponentially in the number of dimensions. That's the multiplicative constant to your big O. For a probabilistic example let x_i ~ N(0,1) for i in 1..d and let s = Sum x_i. Try estimating the probability that -0.1 < s < 0.1 by Monte-Carlo sampling. If most of the probability distribution mass is located near a manifold of lower dimensions, as tends to be the case for natural data, your variance will be huge. MCMC, HMC both improve this state of affair by letting you walk or "glide" on the manifold, but you still have to contend with curvature, with multiple modes etc.

yummyfajitas 10y ago
How does MCMC improve this state of affairs? From what I can see, MCMC only worsens things since you can't be sure you are sampling the entire manifold.
- murbard2 10y ago
  That's a good question. If your distribution isn't multimodal but still suffers from having its mass concentrated in one place, and if the variable you're integrating isn't fat-tailed, then MCMC will help you by sticking around the right area once it finds it.
  - yummyfajitas 10y ago
    I don't even think that's true. Consider a distribution in R^2 that is concentrated on [0,1] x [0,100]. If you have an O(1) jump size it'll take time O(100^2) for the MCMC to fully sample the support of this distribution. If you had a jump size of O(100), then you'll be rejecting 99% of your jumps due to the narrowness in the x-direction. This is hardly an uncommon scenario.
    - murbard2 10y ago
      It's easy to design examples that trip MCMC, but they also trip straight Monte Carlo integration even more. You can, of course, construct examples where MCMC will do worse than simple Monte Carlo integration, but these are uncommon. They mostly illustrate the difficulty of picking an appropriate jump size.
      - yummyfajitas 10y ago
        If you directly draw samples from the underlying distribution, you don't have to worry at all about whether the markov chain or other process (e.g. HMC) is fully sampling the distribution. MC has no rejections and always samples the entire distribution. You simply don't need to worry about trajectories not going everywhere they should.
    - deleted 10y ago
      [deleted]
zump 10y ago
What course did you study to get this knowledge?
- AlexCoventry 10y ago
  Any course on the mathematics of Bayesian inference will introduce you to the necessary tools.
- murbard2 10y ago
  AlexCoventry is right, these are "the mathematics of Bayesian inference", to unpack a bit my recommendation is: - Get a strong grip on linear algebra and euclidean spaces. You should be able to have an intuitive feel for the key theorems (e.g. Cayley-Hamilton, the Spectral theorem) and should be comfortable proving them (at least once). The point isn't that you need to check that they're correct (they are), but if you can prove them, it means you've learned a certain amount of prerequisite knowledge and gained enough familiarity with the topic. They aren't just theorems you apply, they make sense and you understand the intuition behind them. - Get the same feel for multivariate calculus. The intuition there is generally easier to acquire than for linear algebra, but you need to be comfortable with the mechanics of it. Learn to prove your results rigorously, but also to quickly derive formulas by treating infinitesimals as variables, like a physicist. - Study integration, measures, distributions and the fundamentals of probability. If the course talks about sigma-algebras, you're in the right place. Finally study Baseyian statistics, Monte-Carlo Integration, Markov-Chain Monte-Carlo, some information theory You don't need that level of rigor and that level of fundamentals to do machine learning. Some linear algebra, some calculus and some probability theory that you pick along the way will generally do. However, I think if you're interested in ML it is worth the effort because it will make most of the math seamless. This is a lot of math to learn, but it's not particularly "advanced" math. The underlying intuitions are relatively concrete and a lot of the procedures are relatively mechanical.
  - zump 10y ago
    Thanks for the wonderful reply. How long do you estimate it took for you to get to the level you are now?

srean 10y ago

murbard2 and glial have pointed out some caveats. Another thing that ties down Monte Carlo sampling at times is when it is not easy to generate the samples required for Monte Carlo. A common route is to use a Markov chain sampler, hence MCMC, but that's far from a solved problem. Its very hard to reason and test that the Markov chain mixes fast, and if it does not it would take exponential time to generate the samples from its stationary distribution. Coming up with schemes better than a naïve Markov chain is an intellectual industry in itself.

murbard2 10y ago
HMC has better asymptotics than MCMC. If your probability surface is differentiable you should always use that. Riemannian MC takes curvature into account but you need to start storing Hessians which isn't practical. Two missing pieces would be: - An SGD equivalent for sampling "tall" data. The generalized Poisson sampler can do that, but the variance is crappy. - An L-BFGS like version of Riemannian MC, to get rid of the Hessians. Parallel tempering should also be exploited more. That said, at the end of the day, the ideal sampler would be able to reason about the distribution as program and not just as a mathematical black box. It should build tractable approximations intelligently and use those to bootstrap exact sampling schemes. I think we're going to see a wealth of better samplers come out in the next decade, following the path of combinatorial optimization towards preserving the structure of the programs.
- marmaduke 10y ago
  > you need to start storing Hessians aren't there methods for online sparse estimates of the Hessian? I'd expect a lot of "large" problems for which RMC is useful would have sparse structures. > ideal sampler would be able to reason about the distribution as program Are you a developer of Stan? If not, you might be interested. http://mc-stan.org http://mc-stan.org _edit_ by online estimate of a Hessian, I meant online numerical approximation based on the sequence of Jacobians.
  - jmalicki 10y ago
    "An L-BFGS like version of Riemannian MC, to get rid of the Hessians." - that's what he said