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To generate data with a missing at random (MAR) mechanism, usually we can first generate a complete data set and then model the missing probability for the variable $Y$; i.e. $Pr(Y=\text{missing}|{\bf X})$, using a logistical model, i.e. $$Pr(Y_i=1|{\bf X_i})=\frac{\exp(\beta'{\bf X_i})}{1+\exp(\beta'{\bf X_i})}.$$

Based on this probability we can generate the binary variable indicating if $Y$ is missing:

$$I_i = \text{rbinom}(1,p_i).$$

If $I_i=1$ then we delete the corresponding value of $Y$.

If we would like to control the proportion of missing values of $Y$, how can we do that? Since by using the above method the proportion of missing data depends on the values of the $X$'s then the generation of an indicator variable is random, so it's hard to control the overall proportion of missing values.

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The expected number of missing values is the mean of the probabilities

$$f_X(\alpha, \beta) = \frac{1}{n}\sum_i \frac{1}{1 + \exp(-\alpha-X_i\beta^\prime)}$$

where I have written $\alpha$ for the constant term, $\beta$ as a row vector of $d$ coefficients, and followed the convention of arranging the variable values in columns of the design matrix $X$ with $d$ columns (say) and $n$ rows so that $X_i$ is the $i^\text{th}$ row, representing one case.

This is a smooth function of the $d+1$ parameters $\alpha, \beta$ whose range is the full interval $(0,1)$. Therefore, given a target proportion $p$ there exists a $d$-dimensional manifold $f_X^{-1}(p)$ of parameter values for which

$$f_X(\alpha, \beta) =p.$$

I interpret the question as a request for a way to find $\alpha, \beta$ that satisfy this equation.

One way to find a definite solution is to pick a one-dimensional path $t\to\gamma(t)$ in the $(\alpha, \beta)$ space and compute its intersection with $f^{-1}(p)$, which amounts to finding a univariate root. When the path is smooth and intersects $f^{-1}(p)$ transversally, a standard root-finding method such as Newton-Raphson will usually work very well to solve the one-variable equation

$$f(\gamma(t)) - p = 0.$$

For instance, among the infinitely many choices available, you might fix the coefficients $\beta$ and vary $\alpha$, giving the path $t\to (t, \beta)$, or you might fix all the coefficients initially and use the path $t\to (t\alpha, t\beta)$ to scale them uniformly.


As a worked example of the first method (varying only the intercept), consider the following R code. After creating some sample data x and choosing a fixed set of coefficients beta, it sets up a function f to compute $t\to f(\gamma(t))$ and then invokes a root-finding routine uniroot to find zeros of $f(\gamma(t))-p$. Any such zero will be a usable value of $\alpha$, thereby fully specifying the missingness model (which can later be used to generate simulated datasets).

n <- 10; d <- 3               # Set the dimensions of the data
set.seed(17)                  # Create a reproducible starting point
x <- matrix(rnorm(n*d), n, d) # Create independent variables
#
# Determine a model with a given expected proportion of responses 
# missing at random.
#
beta <- 1:d                   # Fix the coefficients at any desired values
f <- function(t) {            # Define a path through parameter space
  sapply(t, function(y) mean(1 / (1 + exp(-y -x %*% beta))))
  # (sapply makes this function vectorizable)
}
#
# Find parameters (alpha, beta) yielding any specified proportions `p`.
#
p <- seq(.1, .9, .2)                        # Specify the target proportions
results <- sapply(p, function(p) {
  alpha <- uniroot(function(t) f(t) - p, c(-1e6, 1e6), tol = .Machine$double.eps^0.5)$root
  c(alpha, f(alpha))})
dimnames(results) <- list(c("alpha", "f(alpha)"), p=p)
print(results)

The output tabulates some solutions for various values of $p$ along with a check of their validity:

          p
                 0.1      0.3       0.5        0.7     0.9
  alpha    -5.102146 -3.45776 -2.251828 -0.7976692 1.27534
  f(alpha)  0.100000  0.30000  0.500000  0.7000000 0.90000

Modify p as desired. Change f and/or beta to search for other sets of parameters.

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  • $\begingroup$ So you meant that the expected proportion of missing values is the MEAN of the probabilities? $f_X(α,β)=\frac{1}{n}∑_i^n\frac{1}{1+exp(−α−X_iβ′)}$. Thank you for your answer, this makes sense. $\endgroup$ – askming Jul 29 '14 at 19:40
  • $\begingroup$ Yes, thank you. Fortunately the code uses the mean as intended. I will change the opening description to match that. $\endgroup$ – whuber Jul 29 '14 at 19:53
  • $\begingroup$ One more thing. In the R code where you define the function f, the sapply wrapper seems not necessary since later on when you use the f function, you only feed it with scalar rather than a vector. Correct me if I'm wrong. Thanks. $\endgroup$ – askming Jul 29 '14 at 20:15
  • $\begingroup$ The vectorization is useful for graphing the function, for using it in an outer product, and all sorts of other things. $\endgroup$ – whuber Jul 29 '14 at 21:26

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