EXERCISE $2$ PLEASE!!!!!!!

In a jupyter notebook named chisquared.ipynb, write some python code to confirm an

assertion made in lecture: Suppose I draw a set of $N$ random variables $\{x_i\}$ from a normal

$($Gaussian$)$ distribution with mean $$ and width $.$ If I define $$ as the weighted sum$-$of$-$

squares deviation from the mean:

$={\displaystyle \underset{i=1}{\overset{N}{}}}(\frac{x_i-}{}{)}^2$

Then $$ will follow a $x^2$ distribution with $k=N$ degrees of freedom.

$(k|)(k|)=\frac{1}{2^{\frac{k}{2}}(\frac{k}{2})}{}^{\frac{k}{2}-1}{e}^{-\frac{}{2}}$

$$ represents the gamma function, but it is easier to generate the $x^2$ PDF directly using

scipy.stats.chi$2.$

For starters, have your code do the following

Set $k=N=10$ data points, $=2\mathrm{.}0$ and $=1\mathrm{.}0.($But note that the $x^2$ distribution

doesn't depend on $$ and $!)$

Perform $1,000($or so$)$ trials in which you sample $NRVs\{x_i\}$ using

numpy.random.normal. Calculate $$ for each trial.

Plot the $$ values in a histogram, using the density$=$True option.

Now, use scipy.stats.chi$2($k$)$ to superimpose the appropriate $x^2$ PDF on your "data" to

confirm they agree.

Here are a few other things to try.

Vary $=2\mathrm{.}0$ and $=1\mathrm{.}0,$ confirm they have no impact on the $x^2$ distribution.

Vary N$,$ confirm that the assertion still holds.

Suppose that, instead of taking $$ as fixed, we determine $$ by taking the mean of the

data points. Since we have determined $$ from the data, the number of degrees of

freedom drops to $k=N-1.$ Recompute and plot the distribution of Q and confirm that

scipy.stats.chi$2($N$-1)$ now describes the data.

Exercise $1.$

In a jupyter notebook named chisquared.ipynb, write some python code to confirm an

assertion made in lecture: Suppose I draw a set of $N$ random variables $\{x_i\}$ from a normal

$($Gaussian$)$ distribution with mean $$ and width $.$ If I define $$ as the weighted sum$-$of$-$

squares deviation from the mean:

$={\displaystyle \underset{i=1}{\overset{N}{}}}(\frac{x_i-}{}{)}^2$

Then $$ will follow a $x^2$ distribution with $k=N$ degrees of freedom.

$(k|)(k|)=\frac{1}{2^{\frac{k}{2}}(\frac{k}{2})}{}^{\frac{k}{2}-1}{e}^{-\frac{}{2}}$

$$ represents the gamma function, but it is easier to generate the $x^2$ PDF directly using

scipy.stats.chi$2.$

For starters, have your code do the following

Set $k=N=10$ data points, $=2\mathrm{.}0$ and $=1\mathrm{.}0.($But note that the $x^2$ distribution

doesn't depend on $$ and $!)$

Perform $1,000($or so$)$ trials in which you sample N RVs $\{x_i\}$ using

numpy.random.normal. Calculate $$ for each trial.

Plot the $$ values in a histogram, using the density$=$True option.

Now, use scipy.stats.chi$2($k$)$ to superimpose the appropriate $x^2$ PDF on your "data" to

confirm they agree.

Here are a few other things to try.

Vary $=2\mathrm{.}0$ and $=1\mathrm{.}0,$ confirm they have no impact on the $x^2$ distribution.

Vary N$,$ confirm that the assertion still holds.

Suppose that, instead of taking $$ as fixed, we determine $$ by taking the mean of the

data points. Since we have determined $$ from the data, the number of degrees of

freedom drops to $k=N-1.$ Recompute and plot the distribution of Q and confirm that

scipy.stats.chi$2(N-1$