 So in this video we're going to look at finding minimal sufficient statistics, and in particular we're going to look at finding the minimal sufficient statistics for mu and lambda of an inverse Gaussian distribution. So recall that the easiest way to find a minimal sufficient statistic is to find a ratio f of xn over f of yn, and these are both the likelihoods, so you have n observations and you treat them as x i's and y, and n observations yi, and you want this to be constant with respect to your parameters of interest. So we say we break this up into two parts, so we have lambda over 2 pi to the power of a half, it's the product from i equals 1 up to n, and well, x to the power of minus 3 over 2, so first of all we'll treat it as a single entire product, and then we'll simplify it out x i minus mu to be squared over 2 mu squared x i, and we do the same with y in terms of y instead of in terms of x in the denominator, so we get lambda over 2 pi to the power of a half yi, that was an x i, to the minus 3 over 2, and the exponential of minus lambda, that's a lambda there, yi minus mu to be squared over 2 mu squared yi. So the first thing we can see, oh and if we close these brackets, the first thing we can see is that these two will cancel immediately, the same in the numerator and the denominator, so you may as well make our life simple, and we say so we have the product i equals 1 to n of x i to the minus 3 over 2 divided by the product i equals 1 to n yi to the minus 3 over 2, that's the first part of it dealt with, and then if I move a product inside an exponential, I get the exponential of the sum, so minus, and I'll take out my common factors to lambda over 2 mu squared, it's the sum for i equals 1 to n of x i squared minus 2 mu x i plus mu squared, and these are all over x i, and we then get the denominator, that's the equivalent in terms of y i, the exponential of minus lambda over 2 mu squared, the sum i equals 1 up to n yi squared minus 2 mu yi plus mu squared all over yi, and close that, again we can simplify it even further, so we get the product i equals 1 up to n of yi over x i, and these are to the, these are to the power of 3 over 2, and then I can bring like terms together, this is going to be come up minus times minus is a plus, so we have the exponential of, and we'll leave the lambda over 2 to outside, so lambda over 2, and we bring, we look at the sum for i equals 1 up to n, and we have a yi squared, so it's going to be, we'll leave the mu squared here for the moment, and we have y i squared divided by y, well that is y i minus 2 mu, and you have y i divided by y i, so that's, they can't see each other out, and plus mu squared over y i, and you do the same with the numerator except for remember all these are going to have reverse sign, so it's going to be the minus x i minus, or plus 2 mu minus mu squared over x i, and then again we simplify this with the product from i equals 1 up to n of y i divided by x i to the power of 3 over 2, and the exponential, and we have lambda over 2 mu squared, and we notice that the minus u and the plus, minus 2 mu plus 2 mu cancel each other out, so you have the sum of i equals 1 up to n of y i minus x i, and then oh look we look in terms of two mu squares on the numerator and the denominator, so we say plus lambda over 2, the mu squared here is going to cancel with here, so lambda over mu, and we have the sum i equals 1 to n, and then we get minus 1 over y i minus 1 over x i, and immediately you can see that this is constant with respect to mu when x i is equal, the sum of x i is equal to y i, so your t of mu, so your your minimal sufficient statistic, so minimal sufficient statistic from mu is when is this constant with respect to mu, the only time mu appears is actually as mu squared, so that's here, so it is where happens when the sum of i equal 1 up to n of x i is equal to the sum of i equal 1 up to n of y i, so t of mu equals the sum of x i, commonly you would treat that as the mean, it's essentially equivalent, the minimal sufficient statistic for lambda, well then you need this term to be zero and this term to be zero for this entire term to be constant with respect to lambda, so you have, you would require it happens when you get the sum of i equals 1 to n of x i equals the sum i equals 1 to n of y i, and that's quite important, the sum i equals 1 to n of x i inverse is equal to the sum i equals 1 to n of y i inverse, so your minimal sufficient statistic for lambda is the sum of x i i equals 1 to n and the sum of x i inverse i equals 1 to n, and they're your minimal sufficient statistics for the parameters of an inverse Gaussian, otherwise known as an inverse normal distribution.