Exercises — Transformations of random variables — change-of-variable technique

What: an affine map $Y=aX+b$ with $a=3,b=-5$. Why the formula applies: $g(x)=3x-5$ is strictly increasing, so it is one-to-one — a single branch. Invert: $x=\frac{y+5}{3}$. Jacobian: $\frac{dx}{dy}=\frac{1}{3}$, magnitude $\frac{1}{3}$. $\boxed{f_Y(y)=\frac{1}{3}{f}_X\left(\frac{y+5}{3}\right).}$ The $\frac{1}{3}$ dilutes the density because stretching the axis by a factor $3$ spreads the same mass over three times the length.

Here $a=\frac{1}{3},b=-\frac{4}{3}$. Using $f_Y(y)=\frac{1}{|a|}{f}_X\left(\frac{y-b}{a}\right)=3f_X(3y+4)$ with the normal density collapses to the standard normal $N(0,1)$: $f_Y(y)=\frac{1}{\sqrt{2\pi }}{e}^{-y^2/2}$. See Normal Distribution. This is exactly standardisation: subtract the mean, divide by the standard deviation.

Monotonic? $g(x)=e^x$ is strictly increasing → one branch. Range: $e^x>0$, so support is $y>0$. Invert: $y=e^x\Rightarrow x=\ln y$. Jacobian: $\frac{dx}{dy}=\frac{1}{y}$ (positive for $y>0$). $f_Y(y)=f_X(\ln y)\cdot \frac{1}{y}=\frac{1}{y\sqrt{2\pi }}\exp \left(-\frac{(\ln y{)}^2}{2}\right),y>0.$ This is the standard log-normal. The extra $\frac{1}{y}$ is the Jacobian, not optional.

$g(x)=\sqrt{x}$ increasing on $(0,1)$; range $y\in (0,1)$. Invert: $x=y^2$. Jacobian $\frac{dx}{dy}=2y$. Since $f_X=1$ on $(0,1)$: $f_Y(y)=1\cdot 2y=2y,0<y<1.$ Check: ${\int }_0^{1}2ydy=1.$ ✓ Density grows toward $y=1$ — square-rooting spreads small $x$ apart and packs large $x$ near $1$.

The figure below plots the fold $g(x)=x^2$ over the support $[-1,2]$. Read it like this: pick a height $y$ on the vertical axis, draw a horizontal line, and count how many times it hits the blue parabola within the dashed lines at $x=-1$ and $x=2$. The yellow line at $y=0.5$ hits it twice (at $x=-\sqrt{y}$ and $x=+\sqrt{y}$, both inside the support), while the red line at $y=2.5$ hits the curve once inside the support (green dot at $+\sqrt{y}$) because the other root $-\sqrt{y}\approx -1.58$ has fallen to the left of $x=-1$ (red ✗, outside). This is the whole reason the density has two different pieces.

• For $0<y<1$: both $+\sqrt{y}$ and $-\sqrt{y}$ lie in $[-1,2]$ (yellow line). Two branches.
• For $1<y<4$: $-\sqrt{y}<-1$ falls outside the support (red ✗), so only $+\sqrt{y}$ survives. One branch.

Jacobian per branch: $\left|\frac{dx}{dy}\right|=\frac{1}{2\sqrt{y}}$, and $f_X=\frac{1}{3}$ wherever defined. $f_Y(y)=\{\begin{array}{ll}\frac{1}{3}\cdot \frac{1}{2\sqrt{y}}\cdot 2=\frac{1}{3\sqrt{y}},& 0<y<1,\\ \frac{1}{3}\cdot \frac{1}{2\sqrt{y}}=\frac{1}{6\sqrt{y}},& 1<y<4,\\ 0,& \text{otherwise.}\end{array}$ Check total mass: ${\int }_0^1\frac{1}{3\sqrt{y}}dy+{\int }_1^4\frac{1}{6\sqrt{y}}dy=\frac{2}{3}+\frac{1}{3}=1.$ ✓

Two roots $\pm \sqrt{y}$, both in the full support $\mathbb{R}$. $f_X$ is even, so both contribute equally: $f_Y(y)=2\cdot \frac{1}{\sqrt{2\pi }}{e}^{-y/2}\cdot \frac{1}{2\sqrt{y}}=\frac{1}{\sqrt{2\pi y}}{e}^{-y/2},y>0.$ This is the chi-square with 1 degree of freedom.

Use the CDF directly (cleaner than densities here): $P(Y\le y)=P(F^{-1}(U)\le y)=P(U\le F(y)).$ The middle step applies $F$ to both sides — legal because $F$ is increasing so it preserves "$\le$." Since $U$ is uniform on $(0,1)$, $P(U\le u)=u$ for $u\in [0,1]$. With $u=F(y)\in [0,1]$: $P(Y\le y)=F(y).$ So $Y$ has exactly CDF $F$. This is why a computer with only a uniform generator can produce any distribution.

Invert the CDF: set $u=1-e^{-\lambda y}\Rightarrow e^{-\lambda y}=1-u\Rightarrow y=-\frac{1}{\lambda }\ln (1-u)$. So $Y=-\frac{1}{\lambda }\ln (1-U)$. Because $1-U$ is also Uniform$(0,1)$, the tidy form $Y=-\frac{1}{\lambda }\ln U$ works too. Density check via change-of-variable ($u=e^{-\lambda y}$, $|du/dy|=\lambda e^{-\lambda y}$, $f_U=1$): $f_Y(y)=\lambda e^{-\lambda y},y>0.$ Exactly $\text{Exp}(\lambda )$.

Invert the linear map: $x=\frac{u+v}{2},y=\frac{u-v}{2}$. Build the Jacobian matrix of the inverse: $J=\left(\begin{array}{cc}\partial x/\partial u& \partial x/\partial v\\ \partial y/\partial u& \partial y/\partial v\end{array}\right)=\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right),\det J=-\frac{1}{2},|\det J|=\frac{1}{2}.$ The determinant is the area-scaling factor of the map — this rotate-and-scale halves areas. $\boxed{f_{U,V}(u,v)=\frac{1}{2}{f}_{X,Y}\left(\frac{u+v}{2},\frac{u-v}{2}\right).}$ Support: the map $(x,y)\mapsto (u,v)$ is a linear bijection of the whole plane, so $(U,V)$ ranges over all of ${\mathbb{R}}^2$ (wherever the transported $f_{X,Y}$ is nonzero); for $X,Y$ supported on $\mathbb{R}$ this is the full ${\mathbb{R}}^2$.

Joint density $f_{X,Y}=\frac{1}{2\pi }{e}^{-(x^2+y^2)/2}$. Substitute the inverse. First expand the exponent, watching the cross-terms: ${\left(\frac{u+v}{2}\right)}^2+{\left(\frac{u-v}{2}\right)}^2=\frac{u^2+2uv+v^2}{4}+\frac{u^2-2uv+v^2}{4}=\frac{u^2+v^2}{2}.$ The $\pm 2uv$ cross-terms cancel exactly — that is what lets $u$ and $v$ separate cleanly. Now integrate out $v$: $f_U(u)={\int }_{-\infty }^{\infty }\frac{1}{2}\cdot \frac{1}{2\pi }{e}^{-(u^2+v^2)/4}dv=\frac{1}{4\pi }{e}^{-u^2/4}\underset{=2\sqrt{\pi }}{\underbrace{{\int }_{-\infty }^{\infty }{e}^{-v^2/4}dv}}.$ Why ${\int }_{-\infty }^{\infty }{e}^{-v^2/4}dv=2\sqrt{\pi }$: use the standard Gaussian integral ${\int }_{-\infty }^{\infty }{e}^{-\alpha x^2}dx=\sqrt{\pi /\alpha }$ with $\alpha =\frac{1}{4}$, giving $\sqrt{\pi /(1/4)}=\sqrt{4\pi }=2\sqrt{\pi }$. Hence $f_U(u)=\frac{1}{4\pi }{e}^{-u^2/4}\cdot 2\sqrt{\pi }=\frac{1}{\sqrt{4\pi }}{e}^{-u^2/4}.$ That is $N(0,2)$ (variance $2$). MGF cross-check: each $N(0,1)$ has $M(t)=e^{t^2/2}$; independence multiplies MGFs, giving $e^{t^2/2}\cdot e^{t^2/2}=e^{t^2}$, the MGF of $N(0,2)$. ✓ (See Expectation and LOTUS for why MGFs multiply.)