Chi-squared distribution  2 N N = number of degrees of freedom Computed using incomplete gamma function: Moments of  2 distribution:

Constructing  2 from Gaussians - 1 Let G(0,1) be a normally-distributed random variable with zero mean and unit variance. For one degree of freedom: This means that: -a a G(0,1) 2121 a2a2 i.e. The  2 distribution with 1 degree of freedom is the same as the distribution of the square of a single normally distributed quantity.

Constructing  2 from Gaussians - 2 For two degrees of freedom: More generally: Example: Target practice! If X 1 and X 2 are normally distributed: i.e. R 2 is distributed as chi-squared with 2 d.o.f X1X1 X2X2

Data points with no error bars If the individual  i are not known, how do we estimate  for the parent distribution? Sample mean: Variance of parent distribution: By analogy, define sample variance: Is this an unbiased estimator, i.e. is =  2 ?

Estimating  2 – 1 Express sample variance as: Use algebra of random variables to determine: Expand: (Don’t worry, all will be revealed...)

Aside: what is Cov(X i,X)? X XiXi

Estimating  2 – 2 We now have For s 2 to be an unbiased estimator for  2, need A=1/(N-1):

If all observations X i have similar errors  : If we don’t know use X instead: In this case we have N-1 degrees of freedom. Recall that: (since =N) Degrees of freedom – 1

Degrees of freedom – 2 Suppose we have just one data point. In this case N=1 and so: Generalising, if we fit N data points with a function A involving M parameters  1...  M : The number of degrees of freedom is N-M.

Example: bias on CCD frames Suppose you want to know whether the zero- exposure (bias) signal of a CCD is uniform over the whole image. First step: determine s 2 (X) over a few sub- regions of the frame. Second step: determine X over the whole frame. Third step: Compute In this case we have fitted a function with one parameter (i.e. the constant X), so M=1 and we expect = N - 1 Use  2 N - 1 distribution to determine probability that  2 >  2 obs