Transformations always give me trouble, but this one does in particular.(adsbygoogle = window.adsbygoogle || []).push({});

Assume [itex]X_1[/itex], [itex]X_2[/itex] independent with binomial distributions of parameters [itex]n_1[/itex], [itex]n_2[/itex], and [itex]p=1/2[/itex] for each.

Show [itex]Y = X_1 - X_2 + n_2[/itex] has a binomial distribution with parameters [itex]n= n_1 + n_2[/itex], [itex]p = 1/2[/itex].

My first instinct was to pick a variable [itex]Z = X_2[/itex], define a joint distribution of [itex]Y[/itex] and [itex]Z[/itex], and sum over all values of [itex]Z[/itex]. I ran into some complex algebra when summing this joint distribution over all values of [itex]Z[/itex], [itex]0[/itex] to [itex]n_2[/itex]. If anyone knows how to sum over all values of z for (n_1 choose y+z-n_2)*(n_2 chooze z) so as to get (n_1 + n_2 choose y), I would love to hear how, but I'm pretty sure this is a no-go.

My next thought was to still choose [itex]Z = X_2[/itex], but this time get the mgf of Y and Z. This boils down to [itex](1/2 + exp(t_1)/2)^{n_1}(1/2 + exp(t_2)/2)^{n_2}[/itex]. When I set [itex]t = t_1 + t_2[/itex], I get an mgf which fits what we're looking for, i.e. the binomial distribution with parameters [itex]n_1[/itex], [itex]n_2[/itex], and [itex]p=1/2[/itex]. But I don't know if that is valid algebra, as a means of obtaining an mgf for a univariate distribution from an mgf for a bivariate distribution.

Any thoughts welcome!

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# Transformation of pmf; bivariate to single-variate

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