Not that this explains why it's getting the wrong answer but maybe it shows how. Consider

z = TransformedDistribution[2 x - 1,  x \[Distributed] BinomialDistribution[1, 1/2]];
PDF[z]

The output is

So when one writes

PDF[z][Range[-2, 2]]

The function is evaluated according to the formula in the output and the formula does not exclude 0 from the range of legitimate values. And plugging zero into that formula gets what you observed for the probability of zero being $2/\pi$. As you noticed that things work fine with a Bernoulli with the resulting function being

which only allows two positive values (and therefore no errors).

Single values seem to work fine in that the check for an appropriate value occurs. But when a list is given, the formula that shows with PDF[z] is used literally.

Mike,

It seems to be a problem of how the PDF function is mapping itself onto a list, rather than a problem with TransformedDistribution. I get the correct answer with:

PDF[z, #] & /@ Range[-2, 2]
{0, 1/2, 0, 1/2, 0}

Which shows that the PDF function itself is fine. I get the wrong answer even if I do:

 PDF[z, {-1, 0}]
{1/2, 2/\[Pi]}

Cannot say that I understand why. The function description states that it is Listable.

Best,

OL.