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Let a stochastic process defines as: $$X(t+1)=A X(t)+B U(t)$$ with: $X(t) \in R^n$, $U(t) \sim N(0,Q_t)$, $Q_t$ semi-positive-definite of size $n \times n$, $X(0) \sim N(0,W_0)$, $A$ of size $n \times n$, $B$ of size $n \times n$.

Is there any condition on $A$ and $B$ to state the existence and the form of the conditional pdf $p_{X_t|X_{t-1}}(x_t|x_{t-1})$?

Thanks in advance.

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Is it homework? – Ilya May 3 '12 at 15:04
Sorry for the typo. This is not homework, just a question from a curious guy. – peuhp May 3 '12 at 15:05
Care to accept answers to your questions? – Did Sep 19 '12 at 18:41

Forget the indices $t$, one is interested in the distribution of $Y=AX+BU$ conditionally on $X$, where the distribution of $X$ is irrelevant and $U$ is independent on $X$ and with centered normal distribution with variance-covariance $Q$. Thus the distribution of $BU$ is centered normal with variance-covariance matrix $C=BQB^*$ and the conditional distribution of $Y$ conditionally on $[X=x]$ is normal with mean $Ax$ and variance-covariance matrix $C$.

Edit The conditional distribution of $Y$ conditionally on $[X=x]$ has a density if and only if it is full-dimensional normal, which happens if and only if $C$ has rank $n$, that is, if and only if $Q$ and $B$ both have rank $n$. Otherwise, for every $x$ in $\mathbb R^n$, the conditional distribution of $Y$ conditionally on $[X=x]$ is concentrated on a hyperplane of $\mathbb R^n$, thus it has no density with respect to the Lebesgue measure on $\mathbb R^n$.

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My question comes from the following note ( page 9): "in general, transition kernel Qk(x, dx′) does not admit a density. Indeed, conditionnally to X_{k−1} = x, r.v. X_k belongs to subset M(x) = {x′ ∈ Rm : there exist w ∈ R^p such that x′ = f_k(x,w)} if p < m and under some mild regularity assumptions, this subset of Rm has zero Lebesgue measure therefore, conditionnally to Xk−1 = x, and the distrib Qk(x, dx′) of r.v. X_k cant have a density" I thought that the linear stochastic process could be affected by this phenomemum, why not? – peuhp May 3 '12 at 15:43
Thanks for your helpful and clear explanation! This maybe extends a bit too much my first question but can you please give me a formal insight of why if "the conditional distribution of Y conditionally on [X=x] is concentrated on a hyperplane of Rn, thus it has no density with respect to the Lebesgue "? Thanks in advance – peuhp May 4 '12 at 6:53
Let $m$ be a measure concentrated on a hyperplane $H$ of $\mathbb R^n$, then $m(\mathbb R^n\setminus H)=0$ and Leb$(H)=0$, hence $m$ and Leb are mutually singular, in particular $m$ has no density with respect to Leb. – Did May 4 '12 at 6:56
Thanks for your help. – peuhp May 7 '12 at 9:53

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