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Let $x_0 \sim N(\mu_0,\Sigma_0)$, where $\mu_0 \in \mathbb R^n$ and $\Sigma_0 \in \mathbb R^{n \times n}$. Then, for $k = 0,\dots,N-1$, let $$x_{k+1} = Ax_k$$ for $A \in \mathbb R^{n \times n}$. Next, define $$E_{x_0,x_1,\dots,x_{N}}[x_0] = \int_{\mathbb R^n} \int_{\mathbb R^n} \cdots \int_{\mathbb R^n} x_0 p(x_0,x_1,\dots,x_N) \text{d}x_N \dots \text{d}x_1 \text{d}x_0$$ and $$E_{x_0}[x_0] = \int_{\mathbb R^n} x_0 p(x_0) \text{d}x_0$$ Is it true that $$E_{x_0,x_1,\dots,x_{N}}[x_0] = E_{x_0}[x_0]$$ It would seem to me that the answer is yes. First, the sequence $x_0 \to x_1 \to \cdots \to x_N$ forms a Markov chain, such that the joint distribution $p(x_0,x_1,\dots,x_N)$ simplifies to $$p(x_0,x_1,\dots,x_N) = p(x_N \mid x_{N-1})p(x_{N-1} \mid x_{N-2})\cdots p(x_1 \mid x_0) p(x_0)$$ Therefore, we can write \begin{align} E_{x_0,x_1,\dots,x_{N-1}}[x_0] &= \int_{\mathbb R^n} \int_{\mathbb R^n} \cdots \int_{\mathbb R^n} x_0 p(x_0,x_1,\dots,x_N) \text{d}x_N \dots \text{d}x_1 \text{d}x_0 \\ &= \int_{\mathbb R^n} \int_{\mathbb R^n} \cdots \int_{\mathbb R^n} x_0 p(x_N \mid x_{N-1})p(x_{N-1} \mid x_{N-2})\cdots p(x_1 \mid x_0) p(x_0) \text{d}x_N \dots \text{d}x_1 \text{d}x_0 \\ &= \int_{\mathbb R^n} \int_{\mathbb R^n} \cdots \int_{\mathbb R^n} p(x_N \mid x_{N-1}) \text{d}x_N \dots p(x_1 \mid x_0) \text{d}x_1 x_0 p(x_0) \text{d}x_0 \end{align} Then, the integrals $$\int_{\mathbb R^n} p(x_N \mid x_{N-1}) \text{d}x_N$$ and $$\int_{\mathbb R^n} p(x_{N-1} \mid x_{N-2}) \text{d}x_{N-1}$$ and so on evaluate to $1$, and so we are left with \begin{align} E_{x_0,x_1,\dots,x_{N-1}}[x_0] &= \int_{\mathbb R^n} \int_{\mathbb R^n} \cdots \int_{\mathbb R^n} p(x_N \mid x_{N-1}) \text{d}x_N \dots p(x_1 \mid x_0) \text{d}x_1 x_0 p(x_0) \text{d}x_0 \\ &= \int_{\mathbb R^n} x_0 p(x_0) \text{d}x_0 \\ &= E_{x_0}[x_0] \end{align} Is my reasoning correct?

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  • $\begingroup$ Why do you think it forms a Markov chain instead of being deterministic? Imagine $A$ is a square matrix of all ones; then at each step, the first element of $x_{i+1}$ is just the sum of the elements of $x_i$, etc. It's a matrix multiplication, not a transition matrix. $\endgroup$
    – jbowman
    Commented Feb 2, 2023 at 0:57
  • $\begingroup$ As written, $x_k = A^kx_0$, so $\mathbb{E}x_k = A^k\mathbb{E}x_0$ $\endgroup$
    – jbowman
    Commented Feb 2, 2023 at 1:04
  • $\begingroup$ @jbowman What I mean by "the sequence $x_0 \to x_1 \to \cdots \to x_N$ forms a Markov chain" is that $x_{k}$ is conditionally independent of $x_{k-2}$ given $x_{k-1}$ for $k = 2,\dots,N-1$. Is there something wrong with saying so? $\endgroup$
    – mhdadk
    Commented Feb 2, 2023 at 14:56
  • $\begingroup$ It's deterministic. One could say that $x_{i+t} = x_i + 1$ is a Markov chain, but why? There's no randomness in the transitions; describing multiplication and addition as a Markov chain adds nothing useful to the discussion. $\endgroup$
    – jbowman
    Commented Feb 2, 2023 at 16:05
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    $\begingroup$ Ah, I see... that's actually kind of clever, if roundabout! $\endgroup$
    – jbowman
    Commented Feb 2, 2023 at 17:55

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You don't need the Markov property. For any two jointly defined random variables $X$ and $Y$, $$E_X[X]=E_{X,Y}[X]$$ The marginal distribution of $X$ is just the distribution of $X$. Now just take $X$ to be $X_0$ and $Y$ to be everything else.

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