# Inequality involving probability of stationary zero-mean Gaussian

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1. Jan 2, 2016

### JohanL

1. The problem statement, all variables and given/known data
Let $$(X(n), n ∈ [1, 2])$$ be a stationary zero-mean Gaussian process with autocorrelation function
$$R_X(0) = 1; R_X(+-1) = \rho$$
for a constant ρ ∈ [−1, 1].
Show that for each x ∈ R it holds that
$$max_{n∈[1,2]} P(X(n) > x) ≤ P (max_{n∈[1,2]} X(n) > x)$$
Are there any values of ρ for which this inequality becomes an equality?
2. Relevant equations

3. The attempt at a solution
$$P(max_{n∈[1,2]} X(n) > x) = P ((X(1) > x) ∪ (X(2) > x)) =$$
$$= P(X(1) > x) + P(X(2) > x) − P ((X(1) > x) ∩ (X(2) > x)) =$$
$$=2 (1 − Φ(x)) − P ((X(1) > x) ∩ (X(2) > x))$$

I cant figure out what to do with the left side of the inequality or how to use the autocorrelation function.

2. Jan 2, 2016

### andrewkirk

The statement of the problem is too unclear to make anything of it. Was the original problem statement the same as above?

Do you mean that there are two Gaussian processes, $\Big(X(1)_t\Big)_{t\in\mathscr{I}_1}$ and $\Big(X(1)_t\Big)_{t\in\mathscr{I}_2}$? If so, what are their index sets $\mathscr{I}_1$ and $\mathscr{I}_2$ (eg are they discrete or continuous processes?), and what is the dependence between the two processes (eg should we assume that they are independent?).

3. Jan 2, 2016

### Ray Vickson

I can't figure out your question: do you have
(1) $n$ continuous, so you have a process $\{ X(t), 1 \leq t \leq 2 \}$ (using $t$ instead of $n$); or do you have
(2) a discrete process $\{ X(n), n \in \{ 1,2 \} \} = \{X_1,X_2 \}$, with $E X_1 = E X_2 = 0$, $E X_1^2 = E X_2^2 = 1$ and $E X_1 X_2 = \rho$?

Judging by your solution attempt, it seems that you mean the latter. So, since $X_1$ and $X_2$ are both standard (unit) normals, albeit correlated, the left-hand-side involves only the marginal distributions, which are identical. Thus,
$$\text{LHS} = \max \{ P(X_1 > x),P(X_2 > x) \} = P(X> x),$$
where $X$ is a unit normal.
As you have indicated, the right-hand-side is
$$\text{RHS} = 2 P(X > x) - P(X_1>x,X_2>x)$$
It is easy enough to examine the two extreme cases $\rho = 0$ (independent) and $\rho = 1$ (essentially, $X_1 = X_2 = X$). For the intermediate cases of $-1 < \rho < 0$ or $0 < \rho < 1$, you might be able to find (in the library, or on-line) the appropriate bounding formulas on $P(X_1 > x, X_2 > x)$, but I don't know of them off-hand.

Last edited: Jan 2, 2016
4. Jan 2, 2016

To show that $$\max\limits_{n\in[1,2]}P(X(n)>x)\leq P(\max\limits_{n\in[1,2]}X(n)>x)$$, I would suggest first showing that $$P(X(n)>x)\leq P(\max\limits_{m\in[1,2]}X(m)>x) \text{ , for all } n\in[1,2]$$. (To avoid confusion, I used the variable m, instead of n, on the right hand side of this inequality.)

We may use basic principles to show that this inequality is true; no knowledge of the autocorrelation function, or even of the process that governs X, is needed.

5. Jan 3, 2016

### JohanL

Thanks all.

Yeah i think you got it right with 2. It was probably an unclear problem statement and maybe my bad latex that made you all a little confused.

I found out that the rest of the solution should be

"$$P ((X(1) > x) ∩ (X(2) > x)) = P(X(1) > x) = P(X(2) > x) = 1 − Φ(x)$$

when ρ = 1

(so that X(1) and X(2) are perfectly positively correlated), giving equality in the desired inequality"

I guess this should be obvious to understand but what definition (equations) implies the above?

", while by conditioning

$$P ((X(1) > x) ∩ (X(2) > x)) < P(X(1) > x) = P(X(2) > x) = 1−Φ(x)$$

for ρ < 1, giving strict inequality in the desired inequality."

Here i am not sure whats going on either. Where is the conditioning.

Last edited: Jan 3, 2016
6. Jan 3, 2016

### Ray Vickson

There is no conditioning going on anywhere in this problem (if, by conditioning you mean the use of conditional probability).

In fact, for any two random variables on the whole line, with identical marginal distributions of mean 0 and variance 1 (whether normal or not, whether dependent, independent, or whatever) we have that your desired inequality is strict for any real x. Properties of the normal distribution are irrelevant to the argument and conclusion: just make a sketch of the regions $\{X_1 > x \: \cap \; X_2 > x \}$ in the $(x_1,x_2)$-plane to see what is happening.

If the random variables are bounded, the inequality still holds but can be non-strict (i.e., an equality) for sufficiently large |x|.

7. Jan 3, 2016

### JohanL

Im probably misunderstanding but...

Maybe I wasnt clear.

I quoted the lecturer's solution:

$$P ((X(1) > x) ∩ (X(2) > x)) = P(X(1) > x) = P(X(2) > x) = 1 − Φ(x)$$

when $\rho = 1$.

(so that X(1) and X(2) are perfectly positively correlated), giving equality in the desired inequality

, while by conditioning

$$P ((X(1) > x) ∩ (X(2) > x)) < P(X(1) > x) = P(X(2) > x) = 1−Φ(x)$$

for ρ < 1, giving strict inequality in the desired inequality.

8. Jan 3, 2016

### andrewkirk

How did you and your lecturer justify this step:
Assuming that $\Phi$ has the usual meaning of the CDF of the standard normal distribution, this statement appears to assume that the variance of every random variable is 1 and, as far as I can see, that is not stated in the problem statement.

9. Jan 3, 2016

### JohanL

I dont know how he justify it or if its correct.

I found it online. Its his solution to an old exam task.

I think its safest i post a picture of it:

10. Jan 3, 2016

### Ray Vickson

The lecturer is doing it the hard way. As I said already in #6 (and 'pizzasky' said first in #4), the result is true in general as long as the marginal distributions of $X_1,X_2$ are identical and are on the whole real line. The random variables $X_1,X_2$ need not be normally-distributed; they need not even have finite means or variances! The only thing that matters is the identical distribution of the marginals. Independence does not matter; correlation does not matter; nothing else matters.