Today we saw how to transform random variables. By this means we can discover further interesting relationships between distributions.
I mentioned in this lecture that if $X,Y$ are independent $N(0,1)$ then $W=X^2+Y^2$ and $T=\tan^{-1}(Y/X)$ are independent $\mathscr{E}(1/2)$ and $U[-\pi/2,\pi/2]$. We can use this result in the reverse direction to generate normal random variables. Make two independent samples from $U[0,1]$, say $U$ and $V$. Let $R^2=-2\log U$ (which is $\mathscr{E}(1/2)$) and let $ \Theta=(V-\frac{1}{2})\pi$. Then $X=R\cos\Theta$ and $Y=R\sin\Theta$ are independent $N(0,1)$.
There is a particularly strange fact in Examples sheet 4, #19.
By the way, if (like me) you have ever been puzzled why it is that the absolute value of the Jacobian, $|J|$, gives the volume change of a $n$-dimensional parallelepiped under an linear transformation in $\mathbb{R}^n$, you might enjoy reading this essay A short thing about determinants, or “an attempt to explain volume forms, determinants and adjugates by staying in 3-d”, by Gareth Taylor.
I have also now added to the notes an Appendix C, in which I explain how under a linear transformation a sphere is mapped into an ellipsoid of a volume that greater by a factor $|J|$. I think that in previous years lecturers have said at this point "you already know this from Vector Calculus". But I am not sure that's that case, and so decided it would be nice to try to remove some of the mystery.
The convolution we need to add two Cauchy random variables is tricky. On page 81, the partial fractions method of computing this integral is to write
$\frac{1}{(1+x^2)(1+(z-x)^2)}=\frac{A+Bx}{(1+x^2)}+\frac{C+D(x-z)}{(1+(z-x)^2)}$
for $(A,B,C,D)=\frac{1}{4+z^2}(1,2/z,1,2/z)$. Upon integrating w.r.t. $x$ the terms with $B$ and $D$ give $0$ (by symmetry either side of $0$), and the terms with $A$ and $C$ provide the claimed result of $\frac{2}{\pi(4+z^2)}$.
I mentioned in this lecture that if $X,Y$ are independent $N(0,1)$ then $W=X^2+Y^2$ and $T=\tan^{-1}(Y/X)$ are independent $\mathscr{E}(1/2)$ and $U[-\pi/2,\pi/2]$. We can use this result in the reverse direction to generate normal random variables. Make two independent samples from $U[0,1]$, say $U$ and $V$. Let $R^2=-2\log U$ (which is $\mathscr{E}(1/2)$) and let $ \Theta=(V-\frac{1}{2})\pi$. Then $X=R\cos\Theta$ and $Y=R\sin\Theta$ are independent $N(0,1)$.
There is a particularly strange fact in Examples sheet 4, #19.
By the way, if (like me) you have ever been puzzled why it is that the absolute value of the Jacobian, $|J|$, gives the volume change of a $n$-dimensional parallelepiped under an linear transformation in $\mathbb{R}^n$, you might enjoy reading this essay A short thing about determinants, or “an attempt to explain volume forms, determinants and adjugates by staying in 3-d”, by Gareth Taylor.
I have also now added to the notes an Appendix C, in which I explain how under a linear transformation a sphere is mapped into an ellipsoid of a volume that greater by a factor $|J|$. I think that in previous years lecturers have said at this point "you already know this from Vector Calculus". But I am not sure that's that case, and so decided it would be nice to try to remove some of the mystery.
The convolution we need to add two Cauchy random variables is tricky. On page 81, the partial fractions method of computing this integral is to write
$\frac{1}{(1+x^2)(1+(z-x)^2)}=\frac{A+Bx}{(1+x^2)}+\frac{C+D(x-z)}{(1+(z-x)^2)}$
for $(A,B,C,D)=\frac{1}{4+z^2}(1,2/z,1,2/z)$. Upon integrating w.r.t. $x$ the terms with $B$ and $D$ give $0$ (by symmetry either side of $0$), and the terms with $A$ and $C$ provide the claimed result of $\frac{2}{\pi(4+z^2)}$.
