Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$X_1$ and $X_2$ are two binary stochastic variables with simultaneous probabilitydistribution:

$$ \begin{matrix} &X_2=0&X_2=1\\ X_1=0&0.30&0.17\\ X_1=1&0.08&0.45 \end{matrix} $$

I want to calculate the correlation between $X_1$ and $X_2$.

Correlation[{{0.30, 0.17}, {0.08, 0.45}}]

The output I get is {{1,-1},{-1,1}}

The solution I seek is: 0.05011

It should be pretty simple, but I have no idea where I am wrong.

share|improve this question
up vote 5 down vote accepted

You could set this up in symbolic form as a bivariate distribution with pmf $f(x,y)$:

Then, using the mathStatica add-on to Mathematica, the correlation you seek is:

Corr[{x, y}, f]


Note that this is slightly different to the solution you posted,as the numerical value is: 0.501123... (not 0.0501).

You can make Mma do this operation too, by itself, as per:

dist = ProbabilityDistribution[f, {x, 0, 1, 1}, {y, 0, 1, 1}]

where f is the piecewise function above, and then evaluate:


The problem with your use of Correlation[{{0.30, 0.17}, {0.08, 0.45}}] is this ... You could use Correlation[xdata, ydata] to find the sample correlation between xdata and ydata ... but (a) you are not seeking a sample correlation ... you are seeking the population correlation, and (b) {{0.30, 0.17}, {0.08, 0.45}} is not your data ... it represents the pmf or distribution of the population data.

share|improve this answer
Great solution. When I try to work futher with the ProbabilityDistribution-function (´dist´), then I cannot use it with Probability[...] I have an example with $X_1$ conditioned $X_1 \neq X_2$. How come? – Jens Jensen Aug 6 '13 at 11:08
Would need to see the exact example to be able to comment on that ... – wolfies Aug 6 '13 at 11:17
How would you calculate: $P(X_1=0|X_1 \neq X_2)$ and $P(X_1=1|X_1 \neq X_2)$. If it is any help. the results are 0.68 and 0.32 – Jens Jensen Aug 6 '13 at 11:22
You could continue: Probability[x == 0 \[Conditioned] x != y, {x, y} \[Distributed] dist] ... and ... Probability[x == 1 \[Conditioned] x != y, {x, y} \[Distributed] dist] .. works fine – wolfies Aug 6 '13 at 11:39
Checked it in v8 and v9 ->> works fine on both on my Mac. Suggest check your input carefully ... or quit kernel and start from fresh :) – wolfies Aug 6 '13 at 12:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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