Transformation of pmf; bivariate to single-variate

[rayge]

Transformations always give me trouble, but this one does in particular.

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

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

My first instinct was to pick a variable $Z = X_2$, define a joint distribution of $Y$ and $Z$, and sum over all values of $Z$. I ran into some complex algebra when summing this joint distribution over all values of $Z$, $0$ to $n_2$. 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 $Z = X_2$, but this time get the mgf of Y and Z. This boils down to $(1/2 + exp(t_1)/2)^{n_1}(1/2 + exp(t_2)/2)^{n_2}$. When I set $t = t_1 + t_2$, I get an mgf which fits what we're looking for, i.e. the binomial distribution with parameters $n_1$, $n_2$, and $p=1/2$. 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!

 

[mathman]

Let Z = n₂ - X₂. Z is binomial - same parameters as X₂. This should be easier. (X₁ + Z).

 

[rayge]

Thanks for the suggestion. From this I get:
$$f(y)=\sum_{z=0}^{n_2} \binom{n_1}{y-z}\binom{n_2}{n_2-z}\Big(\frac{1}{2}\Big)^{n_1+n_2}$$
What I want eventually is this:
$$f(y)=\binom{n_1+n_2}{y}\Big(\frac{1}{2}\Big)^{n_1+n_2}$$
I want very much to snap my fingers and call these equal, but I don't see it. Expanding the sum I get (apologies if I messed this up):
$$\Big(1/2\Big)^{n_1+n_2}\Big(\frac{{n_1!}{n_2!}}{{y!}{(n_1-y)!}{n_2!}{0!}}+\frac{{n_1!}{n_2!}}{{(y-1)!}{(n_1-y+1)!}{(n_2-1)!}{1!}}+\frac{{n_1!}{n_2!}}{{(y-2)!}{(n_1-y+2)!}{(n_2-2)!}{2!}}+\cdots+\frac{{n_1!}{n_2!}}{{(y-n_2)!}{(n_1-y+n_2)!}{n_2!}}\Big)$$
The expansion of what I'm looking for looks like this:
$$\Big(\frac{1}{2}\Big)^{n_1+n_2}\frac{{(n_1+n_2)!}}{{y!}{(n_1+n_2-y)!}}$$
Is there some kind of algebra magic I'm missing to get these to equal each other?

Or maybe you were suggesting using this for the mgf approach? Still not sure about getting $t_1=t_2$ (sorry for the typo before).

 

[mathman]

I believe the correct formula for f(y) should be:

$$f(y)=\sum_{z=0}^{n_2} \binom{n_1}{y-z}\binom{n_2}{z}\Big(\frac{1}{2}\Big)^{n_1+n_2}$$

 

[rayge]

Indeed! Anyone curious about how to get the answer from this, check out the Chu-Vandermonde Identity.

(I ended up using the moment generating function E(e^x_1t-tx_2+tn_2), which was easier than the transformation I had been doing.)