<p>Yes, since <span class="math-container">$\tilde{\theta}$</span> is a vector, to define its distribution one needs a <a href="https://en.wikipedia.org/wiki/Covariance_matrix" rel="nofollow noreferrer">covariance matrix</a>. Here <span class="math-container">$I$</span> is the identity matrix, which means that the noise has a zero-mean normal distribution with standard deviation <span class="math-container">$\sigma$</span>, and different components of this noise are uncorrelated.</p>

CC BY-SA 4.0
2017-10-20T21:26:22.707Z
2019-04-01T13:53:32.710Z
2019-04-01T13:53:32.710Z
Maxim
<p>In the paper <a href="https://arxiv.org/pdf/1706.01905.pdf" rel="nofollow noreferrer">Parameter Space Noise for Exploration</a>, the authors describe the noise that they add to the parameter vector as: </p>

<p><span class="math-container">$$
\tilde{\theta} = \theta + \mathcal{N}(0, \sigma^2I)
$$</span></p>

<p>is <span class="math-container">$I$</span> simply the identity matrix, or am I missing something?</p>

<reinforcement-learning><notation>
What is $I$ in the noise described in the paper "Parameter Space Noise for Exploration"?
user6031125
# What is $I$ in the noise described in the paper "Parameter Space Noise for Exploration"?

(Asked by: username_0 on 2017-10-19)

In the paper [Parameter Space Noise for Exploration](https://arxiv.org/pdf/1706.01905.pdf), the authors describe the noise that they add to the parameter vector as:

$$
\tilde{\theta} = \theta + \mathcal{N}(0, \sigma^2I)
$$

is $I$ simply the identity matrix, or am I missing something?

**Tags:** reinforcement-learning, notation


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## 1 answers

### Answer by username_1 on 2017-10-20. Score: 1 | Accepted answer

Yes, since $\tilde{\theta}$ is a vector, to define its distribution one needs a [covariance matrix](https://en.wikipedia.org/wiki/Covariance_matrix). Here $I$ is the identity matrix, which means that the noise has a zero-mean normal distribution with standard deviation $\sigma$, and different components of this noise are uncorrelated.


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