2025-Q3 – Characteristic Functions, Independence, and Poisson Limits in ℝ³
2025 Probability Prelim Exam (PDF) (link pending)
Problem 3 (verbatim)
Let $(X,Y)\in\mathbb R^{2}$ be a random vector. The characteristic function (c.f.) of $(X,Y)$ is
\(\varphi_{(X,Y)}(s,t)=E(e^{i(sX+tY)}),\qquad (s,t)\in\mathbb R^2.\)
It is known that if the characteristic functions of $(X,Y)$ and $(U,V)$ are identical then they are equal in distribution.
(a)
Let $\varphi_X(s)$ and $\varphi_Y(t)$ be the c.f.’s of $X$ and $Y$.
(i) If $X,Y$ are independent, prove
\(\varphi_{(X,Y)}(s,t)=\varphi_X(s)\,\varphi_Y(t).\)
(ii) If
\(\varphi_{(X,Y)}(s,t)=\varphi_X(s)\,\varphi_Y(t),\)
prove that $X,Y$ are independent.
(b)
Let $X=\sum_{m=1}^n X_m$ in $\mathbb R^3$, where the $X_m$ are i.i.d. and
\(P(X_1=\varepsilon_k)=p_k,\qquad k=1,2,3,\)
with
$\varepsilon_1=(1,0,0)$,
$\varepsilon_2=(0,1,0)$,
$\varepsilon_3=(0,0,0)$, and $\sum p_k=1$.
(i) Find the c.f. of $X_1$, $\varphi_{X_1}(t)=E(e^{i t\cdot X_1})$, and determine how $X_{1,3}$ is distributed.
(ii) Find the c.f. of $X$, $\varphi_X(t)$.
(c)
Let $X_n=\sum_{m=1}^n X_{n,m}$ where $X_{n,m}$ are i.i.d. in $\mathbb R^3$ with
\(P(X_{n,1}=\varepsilon_k)=p_{n,k},\quad k=1,2,3,\)
and assume
\(n p_{n,k} \to \lambda_k,\qquad 0<\lambda_k<\infty,\quad k=1,2,\)
(the condition does not hold for $k=3$).
(i) Show that $\varphi_{X_n}(t)\to \varphi_{X_\infty}(t)$ for some limiting c.f.
(ii) Identify the distributions of the coordinates of $X_\infty$ and describe the relationship between its first two coordinates.
SURVIVAL GUIDE SOLUTION
(a)(i) Independence implies factorization
Claim
If $X$ and $Y$ are independent, then
\(\varphi_{(X,Y)}(s,t)=\varphi_X(s)\varphi_Y(t).\)
Proof
\(\varphi_{(X,Y)}(s,t) =E(e^{i(sX+tY)}) =E\big(e^{isX}e^{itY}\big).\)
By independence of $X$ and $Y$,
\(E(e^{isX}e^{itY})=E(e^{isX})\,E(e^{itY})
=\varphi_X(s)\varphi_Y(t).\)
Conclusion
C.f. of a joint independent pair factorizes.
Key Takeaways
- Independence of random variables implies independence of measurable transforms such as $e^{isX}$.
- Characteristic functions encode independence via multiplicative factorization.
(a)(ii) Factorization implies independence
Claim
If
\(\varphi_{(X,Y)}(s,t)=\varphi_X(s)\varphi_Y(t)\quad\forall (s,t),\)
then $X$ and $Y$ are independent.
Proof
Construct auxiliary random variables $U,V$ with
\(U\stackrel{d}{=}X,\qquad V\stackrel{d}{=}Y,\)
and require $U$ and $V$ to be independent. Their joint c.f. is
\(\varphi_{(U,V)}(s,t)=\varphi_U(s)\varphi_V(t)=\varphi_X(s)\varphi_Y(t).\)
Given the hypothesis,
\(\varphi_{(X,Y)}(s,t)=\varphi_{(U,V)}(s,t).\)
By uniqueness of characteristic functions,
\((X,Y)\stackrel{d}{=}(U,V).\)
Since $(U,V)$ is independent, all of its finite-dimensional distributions factorize, hence
\(P(X\in A, Y\in B)=P(U\in A,V\in B)=P(U\in A)P(V\in B)=P(X\in A)P(Y\in B).\)
Thus $X$ and $Y$ are independent.
Conclusion
Factorization of the joint c.f. is equivalent to independence.
Key Takeaways
- Independence can be transferred through equality in distribution.
- Characteristic functions uniquely determine distribution.
- Proving independence often reduces to comparing c.f.’s.
(b)(i) c.f. of a discrete 3-vector taking only basis vectors
Claim
For $t=(t_1,t_2,t_3)\in\mathbb R^3$, \(\varphi_{X_1}(t)=p_1 e^{it_1}+p_2 e^{it_2}+p_3 e^{it_3}.\)
Proof
\(\varphi_{X_1}(t) =E(e^{i t\cdot X_1}) =\sum_{k=1}^{3} p_k\, e^{i(t\cdot\varepsilon_k)}.\)
Dot products:
\(t\cdot \varepsilon_1=t_1,\qquad
t\cdot \varepsilon_2=t_2,\qquad
t\cdot \varepsilon_3=t_3.\)
Thus
\(\varphi_{X_1}(t)=p_1e^{it_1}+p_2e^{it_2}+p_3e^{it_3}.\)
Distribution of $X_{1,3}$:
Since $\varepsilon_1$ and $\varepsilon_2$ have third coordinate $0$,
\(X_{1,3}=0 \quad\text{with probability } 1.\)
Conclusion
$\varphi_{X_1}(t)$ is a 3-point discrete Fourier transform; third coordinate is degenerate at 0.
Key Takeaways
- For discrete-valued vectors, the c.f. is a weighted exponential sum.
- Degeneracy of coordinates is immediately visible from support vectors.
(b)(ii) c.f. of the sum of i.i.d. vectors
Claim
\(\varphi_X(t)=\big(p_1 e^{it_1}+p_2 e^{it_2}+p_3 e^{it_3}\big)^n.\)
Proof
Because the $X_m$ are independent and identically distributed: \(\varphi_X(t)=\prod_{m=1}^n \varphi_{X_m}(t) = \big(\varphi_{X_1}(t)\big)^n.\)
Conclusion
\(\boxed{\varphi_X(t)=(p_1 e^{it_1}+p_2 e^{it_2}+p_3 e^{it_3})^n.}\)
Key Takeaways
- c.f. of a sum of independent vectors = product of their c.f.’s.
- This is a multinomial-type generating function.
(c)(i) Limit of c.f.’s for triangular array with vanishing probabilities
Claim
\(\varphi_{X_n}(t) =\big(p_{n,1} e^{it_1}+p_{n,2} e^{it_2}+p_{n,3} e^{it_3}\big)^n \;\longrightarrow\; \exp\!\big(\lambda_1(e^{it_1}-1)+\lambda_2(e^{it_2}-1)\big).\)
Proof
From (b)(ii), \(\varphi_{X_n}(t) =\left(1 + (e^{it_1}-1)p_{n,1} + (e^{it_2}-1)p_{n,2}\right)^n,\) because $e^{it_3}=1$ and $p_{n,3}=1-p_{n,1}-p_{n,2}$.
Use the limit: \(\big(1+\tfrac{a_n}{n}\big)^n \to e^{\lim a_n}\) provided $n p_{n,k}\to \lambda_k$.
Rewrite: \(\varphi_{X_n}(t) =\left(1 + \frac{n p_{n,1}}{n}(e^{it_1}-1) + \frac{n p_{n,2}}{n}(e^{it_2}-1)\right)^n.\)
Letting $n\to\infty$, \(\varphi_{X_n}(t) \to \exp\big(\lambda_1(e^{it_1}-1)+\lambda_2(e^{it_2}-1)\big).\)
This is the c.f. of a pair of independent Poisson variables in coordinates 1 and 2.
Conclusion
A limiting random vector $X_\infty$ exists with c.f. \(\varphi_{X_\infty}(t)=\exp\big(\lambda_1(e^{it_1}-1)+\lambda_2(e^{it_2}-1)\big).\)
Key Takeaways
- Vanishing probabilities in triangular arrays produce Poisson limits.
- Third coordinate disappears because its corresponding probability does not scale as $1/n$.
- This is a multivariate version of the classical Poisson limit theorem.
(c)(ii) Distribution and relationships of coordinates of $X_\infty$
Claim
- $X_{\infty,1}\sim\text{Poisson}(\lambda_1)$
- $X_{\infty,2}\sim\text{Poisson}(\lambda_2)$
- $X_{\infty,3}=0$ almost surely
- $X_{\infty,1}$ and $X_{\infty,2}$ are independent
Proof
-
Independence of first two coordinates.
The limiting c.f. factors: \(\varphi_{X_\infty}(t) = \exp(\lambda_1(e^{it_1}-1))\; \exp(\lambda_2(e^{it_2}-1)).\) By part (a)(ii), factorization of a joint c.f. implies independence. -
Marginal distributions.
The c.f.
\(\exp(\lambda_k(e^{it_k}-1))\) is exactly the c.f. of $\text{Poisson}(\lambda_k)$. -
Third coordinate.
Because every $X_{n,1}$ third coordinate is always $0$, all sums satisfy $X_{n,3}=0$.
Therefore the limit must satisfy
\(X_{\infty,3}=0 \quad\text{a.s.}\)
Conclusion
\(X_\infty = (\text{Poisson}(\lambda_1),\text{Poisson}(\lambda_2),0),\) with the Poisson components independent.
Key Takeaways
- Independence in the limit follows from factorization of the limiting c.f.
- Poisson limits appear when $np_{n,k}\to\lambda_k$.
- Degenerate coordinates remain degenerate in the limit.
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