Exercises — Independence of random variables — formal definition

Step 1 — check the support. WHY: independence needs a product (rectangular) support. Here the support is $[0,1]\times [0,1]$, a genuine rectangle. ✅ Step 2 — try to split the formula. $6x^{2}y=(6x^2)\cdot (y)$, or cleaner $=(3x^2)(2y)$. It is a function of $x$ times a function of $y$. ✅ Step 3 — conclude and read off marginals. By the factorisation shortcut, $X\perp Y$. Normalise each piece to a density: $f_X(x)={\int }_0^{1}6x^{2}ydy=6x^2\cdot \frac{1}{2}=3x^2,f_Y(y)={\int }_0^{1}6x^{2}ydx=2y\cdot \frac{1}{2}\cdot {\int }_0^{1}6x^{2}dx\Rightarrow f_Y(y)=2y.$ Check: $f_X{f}_Y=3x^2\cdot 2y=6x^{2}y=f_{X,Y}$. ✅ Independent, with $f_X(x)=3x^2$, $f_Y(y)=2y$.

Step 1 — marginals. $p_X(0)=0.1+0.2=0.3$, $p_X(1)=0.7$; $p_Y(0)=0.4$, $p_Y(1)=0.6$. Step 2 — test cell $(0,0)$. $p_X(0)p_Y(0)=0.3\cdot 0.4=0.12$, but the table says $0.1$. $0.12\ne 0.1$. ✗ Conclusion: one failing cell is enough — NOT independent.

Step 1 — normalise. WHY: a density must integrate to 1. ${\int }_0^{\infty }{\int }_0^{\infty }c{e}^{-2x}{e}^{-3y}dxdy=c\cdot \frac{1}{2}\cdot \frac{1}{3}=\frac{c}{6}=1\Rightarrow c=6.$ Step 2 — factor. $6e^{-2x}{e}^{-3y}=(2e^{-2x})(3e^{-3y})$ on the rectangle $(0,\infty {)}^2$. ✅ Independent. Step 3 — marginals. $f_X(x)=2e^{-2x}$ (Exp rate 2), $f_Y(y)=3e^{-3y}$ (Exp rate 3). Product equals the joint. ✅

Step 1 — product of expectations. WHY: independence lets $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]$. $\mathbb{E}[XY]=2\cdot (-1)=-2.$ Step 2 — variance of a sum. In general $\mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)+2\mathrm{Cov}(X,Y)$. Independence ⇒ $\mathrm{Cov}=0$. $\mathrm{Var}(X+Y)=1+4+0=5.$ (See Covariance and correlation for the covariance term.)

Step 1 — look at the support (figure). WHY first: the shortcut theorem requires a rectangular support. The shaded region is a triangle ($x\le y$), not a box. Already suspicious: knowing $X=0.8$ forces $Y\ge 0.8$ — information leaks. Step 2 — compute marginals honestly. For fixed $x$, $y$ ranges over $[x,1]$: $f_X(x)={\int }_x^{1}8xydy=8x\cdot \frac{1-x^2}{2}=4x(1-x^2),0\le x\le 1.$ For fixed $y$, $x$ ranges over $[0,y]$: $f_Y(y)={\int }_0^{y}8xydx=8y\cdot \frac{y^2}{2}=4y^3,0\le y\le 1.$ Step 3 — compare. $f_X(x)f_Y(y)=4x(1-x^2)\cdot 4y^3=16x{y}^3(1-x^2)$, which is not equal to $8xy$. ✗ Conclusion: NOT independent. The separable-looking formula was a decoy; the triangular support did the coupling.

Step 1 — enumerate. The four $(X,Y)$ points are $(1,0),(0,1),(-1,0),(0,-1)$, each with probability $\frac{1}{4}$. Step 2 — means. $\mathbb{E}[X]=\frac{1}{4}(1+0-1+0)=0$; $\mathbb{E}[Y]=\frac{1}{4}(0+1+0-1)=0$. Step 3 — $\mathbb{E}[XY]$. At each of the four points, one coordinate is $0$, so $XY=0$ always. Thus $\mathbb{E}[XY]=0$ and $\mathrm{Cov}=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]=0$. Uncorrelated. ✅ Step 4 — dependence. $P(X=1)=\frac{1}{4}$ and $P(Y=0)=\frac{1}{2}$, so if independent $P(X=1,Y=0)$ would be $\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8}$. But the actual value is $\frac{1}{4}$ (the point $(1,0)$). $\frac{1}{4}\ne \frac{1}{8}$. ✗ Conclusion: zero covariance, yet dependent — covariance only detects linear structure.

Step 1 — why convolution. For independent $X,Y$, the density of $Z=X+Y$ is the convolution (see Sums of independent random variables — convolution): $f_Z(z)={\int }_{-\infty }^{\infty }{f}_X(t)f_Y(z-t)dt.$ Independence is exactly what lets us multiply $f_X(t)f_Y(z-t)$ inside the integral. Step 2 — plug in. $f_X=f_Y=1$ on $[0,1]$, else $0$. The integrand is $1$ only when $0\le t\le 1$ and $0\le z-t\le 1$, i.e. $\max (0,z-1)\le t\le \min (1,z)$. Step 3 — case split (all cases!).

• $z<0$ or $z>2$: no overlap, $f_Z=0$.
• $0\le z\le 1$: $t\in [0,z]$, length $z$, so $f_Z(z)=z$.
• $1\le z\le 2$: $t\in [z-1,1]$, length $2-z$, so $f_Z(z)=2-z$. This is the triangular density peaking at $z=1$ (figure). Step 4 — evaluate. $f_Z(0.5)=0.5$; $f_Z(1)=2-1=1$ (the peak).

Step 1 — discrete convolution. $P(X+Y=n)={\sum }_{k=0}^{n}P(X=k)P(Y=n-k)$ using independence. Step 2 — the classic identity. Independent Poissons add: $X+Y\sim \text{Poisson}(\lambda +\mu )=\text{Poisson}(5)$. (Proof: the sum ${\sum }_k\frac{e^{-\lambda }{\lambda }^k}{k!}\frac{e^{-\mu }{\mu }^{n-k}}{(n-k)!}=e^{-(\lambda +\mu )}\frac{(\lambda +\mu {)}^n}{n!}$ by the binomial theorem.) Step 3 — read off. $\mathbb{E}[X+Y]=\lambda +\mu =5$. $P(X+Y=0)=e^{-5}\approx 0.006738$.

Goal. Show $P(g(X)\in C,h(Y)\in D)=P(g(X)\in C)P(h(Y)\in D)$ for all Borel $C,D$. Step 1 — pull back the sets. WHY: convert questions about $g(X)$ into questions about $X$. Let $A=g^{-1}(C)=\{x:g(x)\in C\}$ and $B=h^{-1}(D)$. These are Borel sets because $g,h$ are measurable. Then $\{g(X)\in C\}=\{X\in A\},\{h(Y)\in D\}=\{Y\in B\}.$ Step 2 — apply the definition of $X\perp Y$. Independence holds for every pair of Borel sets, in particular for $A,B$: $P(X\in A,Y\in B)=P(X\in A)P(Y\in B).$ Step 3 — translate back. $P(g(X)\in C,h(Y)\in D)=P(X\in A,Y\in B)=P(X\in A)P(Y\in B)=P(g(X)\in C)P(h(Y)\in D).$ Since $C,D$ were arbitrary, $g(X)\perp h(Y)$. $■$ Insight: independence is preserved under any per-variable transformation — you can't create dependence by relabelling each variable separately.

Step 1 — the idea. WHY not just any function: we need it to integrate to 1, stay non-negative, have uniform marginals, yet couple $x$ and $y$. A perturbation trick works: $f_{X,Y}(x,y)=1+\alpha (2x-1)(2y-1),(x,y)\in [0,1{]}^2,$ with $\alpha =\frac{1}{2}$ (any $|\alpha |\le 1$ keeps it $\ge 0$, since $|(2x-1)(2y-1)|\le 1$). Step 2 — non-negativity. With $\alpha =\frac{1}{2}$: minimum value is $1-\frac{1}{2}=\frac{1}{2}>0$. ✅ Step 3 — marginal of $X$. Integrate out $y$; note ${\int }_0^1(2y-1)dy=0$: $f_X(x)={\int }_0^1[1+\alpha (2x-1)(2y-1)]dy=1+\alpha (2x-1)\cdot 0=1.$ So $X\sim U[0,1]$. By symmetry $f_Y(y)=1$, $Y\sim U[0,1]$. ✅ Step 4 — dependence. If independent, we'd need $f_{X,Y}=f_X{f}_Y=1$ everywhere. But at $(x,y)=(1,1)$: $f_{X,Y}=1+\frac{1}{2}(1)(1)=1.5\ne 1$. ✗ Conclusion: uniform marginals, rectangular support, yet dependent — because the joint refuses to factor even though each margin looks innocent.

Every problem above is one question wearing different clothes: does the joint really equal the product of the marginals, at every point, over a rectangular region? If yes — independent, and life is easy (multiply, add variances, convolve sums). If the support is bent (a triangle, a circle) or the formula won't split, they gossip.