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Let $H$ be a separable Hilbert space and $N(0, C)$ and $N(0, D)$ be Gaussian measures on it. Further, for each $v \in H$, define $R_v = \frac{\left\langle v,Cv \right\rangle}{\left\langle v,Dv \right\rangle}$. Basically, $R_v$ is the ratio of Covariance of one dimensional Gaussian measures induced by $N(0, C)$ and $N(0, D)$ along $v$ respectively. Are the following statements equivalent-

1) $N(0, C)$ and $N(0, D )$ are equivalent to each other.

2) $ 0 < \inf_{v \in H} R_v \leq \sup_{v \in H} R_v < \infty.$

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  • $\begingroup$ Define $\mu$ and $\nu$. $\endgroup$
    – mpiktas
    Commented Jan 20, 2015 at 13:52
  • $\begingroup$ Sorry, corrected $\mu$ and $\nu$ were nothing but $N(0, C)$ and $N(0, D)$ respectively. $\endgroup$
    – Madhuresh
    Commented Jan 20, 2015 at 14:27
  • $\begingroup$ No, I don't think this is true. For instance, $N(0,C)$ and $N(0, 2C)$ are typically mutually singular. $\endgroup$
    – Nate Eldredge
    Commented Jan 20, 2015 at 15:40

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They are not equivalent.

For an explicit counterexample, let $\{e_1, e_2, \dots\}$ be an orthonormal basis for $H$, and let $C$ be the diagonal operator $C e_n = \frac{1}{n^2} e_n$. Let $D =2C$. Then $R_v = 1/2$ for every $v$ so (2) is satisfied.

Define random variables $X_n$ on $H$ by $X_n(v) = {n} \langle v, e_n \rangle$. Then under $N(0,C)$, the $X_n$ are iid $N(0,1)$, and under $N(0,D)$ they are iid $N(0,2)$. So by the strong law of large numbers, we have $\frac{1}{n} (X_1(v)^2 + \dots + X_n(v)^2) \to 1$ for $N(0,C)$-a.e. $v$, but likewise we have $\frac{1}{n} (X_1(v)^2 + \dots + X_n(v)^2) \to 2$ for $N(0,D)$-a.e. $v$. So $N(0,C)$ and $N(0,D)$ are mutually singular and (1) fails.

In H. H. Kuo's book Gaussian Measures in Banach Spaces, section II.3, you can find the following theorem:

Theorem. $N(0,C)$ is equivalent to $N(0,D)$ iff there exists a bounded operator $T$ which is positive definite and invertible and such that $I-T$ is Hilbert–Schmidt, with $D = \sqrt{C} T \sqrt{C}$. Otherwise, $N(0,C)$ and $N(0,D)$ are mutually singular.

(Kuo credits this result to Feldman–Hajek, but the reference given is to Varadhan's Stochastic Processes.)

This shows that we do have (1) implies (2), since if $D = \sqrt{C} T \sqrt{C}$ we can write $R_v = \frac{\|\sqrt{C} v\|^2}{\|\sqrt{T} \sqrt{C} v\|^2}$, and from that you can conclude $\|T\|^{-1} \le R_v \le \|T^{-1}\|$.

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  • $\begingroup$ You probably meant to write (1) implies (2)... $\endgroup$
    – Martin Hairer
    Commented Jan 20, 2015 at 17:48
  • $\begingroup$ @MartinHairer: Oops, thank you. Fixed it. $\endgroup$
    – Nate Eldredge
    Commented Jan 20, 2015 at 17:58
  • $\begingroup$ @Nate: Maybe I missed something but the $C$ in your example should be of trace class? $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jan 21, 2015 at 20:51
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    $\begingroup$ @AbdelmalekAbdesselam: Yes, you're right. I got a square root backward in my head somewhere. Fixed. $\endgroup$
    – Nate Eldredge
    Commented Jan 21, 2015 at 21:30
  • $\begingroup$ Since this is a very related question, I thought I would put it here. Assume condition (2) holds. $ k_1 = \inf_{v \in H} R_v \leq \sup_{v \in H} R_v = k_2.$ Will this imply that for balls of small enough radius, the number $N(0,C)(B)$ is sandwitched between $N(0,r_1D)(B)$ and $N(0,r_2D)(B)$ where $B$ is a ball of small enough radius and $r_1$ and $r_2$ are constants depending on $K_1$ and $K_2$ respectively. $\endgroup$
    – Madhuresh
    Commented Feb 7, 2015 at 8:06

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