2
$\begingroup$

I have two distributions, $X$ and $Y$ (shown on the horizontal X axis and vertical Y axis, respectively, see image), that represent different ways of scoring some complex system. For a subset of states of a complex system, I have calculated both a score $x$ using the $X$ method, and a score $y$ using the $Y$ method (i.e. the methods are independent), which produces the plot below. Some more points:

  • The method for scoring X is quick but can be inaccurate
  • The method for scoring Y is slow to compute but extremely robust
  • Larger scores are better
  • Each distribution has a different scale (the distributions are correlated).

I would like to use the $X$ method as a proxy for the $Y$ method. Is it possible to say that some score $x_{cut}$ in $X$, will map to a score $Y \ge 4$, so for any $x \ge x_{cut}$, we will be certain that the corresponding $y \ge 4$ , with a confidence of say 99.95%? If so, how would I calculate $x_{cut}$?

My naive thinking is to choose the largest $x$ value found for all $y \lt 4$, and use that as my cut-off. But as I've only measured scores for some states in my complex system, I realise that I may not have encountered the largest $x$ that corresponds to $y \lt 4$.

y-scores vs x-scores

$\endgroup$
7
  • $\begingroup$ Confidence levels relate to the probability that a certain procedure gives a correct hypothesis. The hypothesis "some score X will give a score Y>=4" is too vague. Could you explain this a bit more. $\endgroup$ Mar 13, 2018 at 9:54
  • $\begingroup$ How one would go about this analysis ought to depend (very strongly) on how you obtained the data. There can be a huge difference in results between, say (1) sampling $(X,Y)$ randomly from a population and (2) specifying a set of $x_i$ and observing the value of $Y$ for each $x_i$. $\endgroup$
    – whuber
    Mar 13, 2018 at 14:35
  • $\begingroup$ @MartijnWeterings I'm looking for a cut-off in X that will ensure that all of the corresponding values in Y are >= 4. I've reworded paragraph 2 to clarify this. I've sampled a subset of the entire population, so if I were to use my naive approach (described in paragraph 3) there would be a confidence level associated with using that value? $\endgroup$ Mar 14, 2018 at 1:48
  • $\begingroup$ @whuber X and Y are independent observations of some complex system. I've updated paragraph 1 to clarify this. $\endgroup$ Mar 14, 2018 at 2:00
  • $\begingroup$ @ilikeprimenumbers As I understand your set-up, there is no way of having a $x_{cut}$ which with 100% certainty assures that e.g. $Y>=4$. So maybe what you asking is something like: Find the $x_{cut}$ so $P(Y>=4|X=x_{cut}) = 90\%$? If yes, this can be derived from the conditional quantile function. This quantile function can be found with Quantile regression, which seems like overkill. I understand that my answer below doesn't exactly answer the quantile question. $\endgroup$
    – Duffau
    Mar 14, 2018 at 10:32

1 Answer 1

0
$\begingroup$

If you are willing to assume the x value is normally distributed you can use Bayes' theorem to calculate the probability of some y score given a value of x.

Bayes' theorem where $X$ is continuous and $Y$ is discrete is given by,

$P(Y=k|X=x) = \frac{f_{X|Y=k}(x)\cdot P(Y=k)}{f_X(x)}$

where $P(Y=k|X=x)$ is the probability of $Y$ is equal to some score $k$, given the x-score is equal to some value $x$. The density function $f_{X|Y=k}$ is the density of $X$ conditional on the y-score. For this density we can use a density, which is normal conditional on $Y$. The numerator is the "global" density of $X$, also we you can assume a normal distribution.

Given some data it is quite easy estimating the different "global" or unconditional probabilities of the y-score ($P(Y=k)$), it's simply the proportions of the scores in the categories 1,2,3,4 and 5.

Also the denominator is quite easy to estimate. The normal distribution is fully characterized by the parameters $\mu$ and $\sigma$, which are the mean and standard deviation, respectively. The empirical counterparts are just the average and the standard deviation of the x-score.

The tricky part is the conditional normal in the numerator. Here you need a set of means, and standard deviations, one for each value of y (1,2,3,4 and 5). So, you need to group the x-values for each corresponding y-value and compute the mean and standard deviation. These values needs to be saved and you can use them to compute the conditional normal for each value of $Y$.

R-code for estimating and computing the conditional probability function

Here is an R example simulating some values and doing the estimation just described. The function conditional_prob computes the probability of some y-score given a x-score. If you want the probability of say Y>=4 given some x-value you can simply sum the probabilities for y=4 and y=5.

# Simulating x and y scores
n <- 100
mean_x <- 35
sd_x <- 2

x <- rnorm(n, mean=mean_x, sd=sd_x)
y <- as.numeric(cut(x, breaks=5))

plot(x,y)

# Estimating parameters for conditional probability
x_grouped_means <- aggregate(x, list(y), mean)
colnames(x_grouped_means) <- c('y', 'mu_hat')

x_grouped_stds <- aggregate(x, list(y), sd)
colnames(x_grouped_stds) <- c('y', 'sigma_hat')

x_global_mean <- mean(x)
x_gloabal_std <- sd(x)

y_relative_freqs <- as.data.frame(table(factor(y))/length(y))
colnames(y_relative_freqs) <- c('y', 'p_hat')


conditional_prob <- function(x, y, grouped_mu_hat_x, grouped_sigma_hat_x, mu_hat_x, sigma_hat_x, p_hat_y) {
  mu_hat_x_given_y <- grouped_mu_hat_x[grouped_mu_hat_x['y']==y, 'mu_hat']
  sigma_hat_x_given_y <- grouped_sigma_hat_x[grouped_sigma_hat_x['y']==y, 'sigma_hat']
  prob_y <- p_hat_y[p_hat_y['y']==y, 'p_hat']
  dnorm(x, mean=mu_hat_x, sd=sigma_hat_x)*prob_y/dnorm(x, mean=mu_hat_x, sd=sigma_hat_x)
}

Compution the probability of Y=>4 given x=35

p4 <- conditional_prob(35, 4, x_grouped_means, x_grouped_stds, x_global_mean, x_gloabal_std, y_relative_freqs)
p5 <- conditional_prob(35, 5, x_grouped_means, x_grouped_stds, x_global_mean, x_gloabal_std, y_relative_freqs)
p4+p5
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.