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The Item Count Technique (ITC) (aka Unmatched Count Technique) is a survey technique to ask a sensitive question while taking account to the anonymity of the respondent's answer. Therefore in standard ITC the sample is divided into two halves $n_1,\ n_2$ from which the respondents are asked different lists of items: a long list and a short list, whereof the long list $Y$ includes the short list $X$ plus the sensitive item. By subtracting the means of the two lists we can derive the estimate of the probability of the sensitive item $\hat\pi_s$ in the population:

$\hat\pi_s=\bar{Y}-\bar{X}.$

The variance is given by

$\mathrm{Var}(\pi_s)=\frac{\pi(1-\pi)}{n_2}+\frac{n\sum\limits_{j=1}^g{\theta_j\Big(1-\sum\limits_{j=1}^g\theta_j\big)}}{n_1n_2}+\frac{n\sum\limits_{k=1;\ j\neq k}^g\theta_j\theta_k}{n_1n_2}.$ (Hussain, Shah & Shabbir 2012)

As the sample is divided in two halfs, it is obvious that the statistical power of standard ITC is lowered accordingly. Therefore there exist several improvements of the standard ITC in order to enhance statistical power, e.g. the ITC double list. The ITC double list has two long lists $Y_1,\ Y_2$ and two short lists $X_1,\ X_2$ and the respondents of each sample halves $n_1,\ n_2$ are asked the two lists crosswise, e.g. $n_1\text:\ Y_1, X_2, \ n_2\text:\ Y2,\ X_1$. According to standard ICT the estimate $\hat\pi_d$ of ICT double list can be derived by

$\hat\pi_{1} = \bar{Y}_1 - \bar{X}_1,\ \hat\pi_{2} = \bar{Y}_2 - \bar{X}_2\quad\Rightarrow \quad\hat\pi_d = \dfrac{\hat\pi_{1} + \hat\pi_{2}}{2}.$

Now for a term paper I wanted to cite the formula for the variance of the ITC double list $\mathrm{Var(\pi_d)}$ as well but it seems I can't find it in any paper though I was searching for hours.

Does anyone know or else could anyone show me how the variance $\mathrm{Var(\pi_d)}$ of the ITC double list is calculated?

Edit:

I only could find following formula, but this but I need the version as seen above for$\mathrm{Var(\pi_s)}$ since I can't calculate the covariance without knowing the variances of each list.

$\mathrm{Var(\pi_d)}=\frac14\Big(\mathrm{Var(\pi_1)}+\mathrm{Var(\pi_2)}+2\mathrm{Cov(\pi_1,\ \pi_2)}\Big)$

jay.sf
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  • *ITC* is new to me but similar older survey-based approaches to obtaining answers to sensitive questions have infinite variance. This may be the reason you can't find its calculation in the literature. What would have been nice is that the authors clearly stated that the variance was infinite. Why not ask them directly about it? – Mike Hunter Jan 28 '18 at 14:21
  • @DJohnson Perhaps we're talking of different variances? How could there be an infinite variance in a finite sample? [Wikipedia: Unmatched count](https://en.wikipedia.org/wiki/Unmatched_count) – jay.sf Jan 28 '18 at 14:36
  • Undefined or infinite variances are ubiquitous in finite data samples. E.g., this thread discusses the issue... https://stats.stackexchange.com/questions/91512/how-can-a-distribution-have-infinite-mean-and-variance Your case is not concerned with the distribution of the information but with the method involved in gathering it. – Mike Hunter Jan 28 '18 at 15:46

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