I am trying to evaluate E [ sign ⁡ ⟨ v , z ⟩ ] for v ∈ R n fixed and z i ∼ N ( 0 , 1 ) ∀ i ∈ [ n ]The sign part is what is confusing me. Clearly, E [ sign ⁡ ⟨ v , z ⟩ ] = E [ sign ⁡ ∑ i = 1 n v i z i ] = P [ ∑ i = 1 n v i z i &gt; 0 ] − P [ ∑ i = 1 n v i z i &lt; 0 ] .But I don't know how to simplify further (i.e. to get the expectation just on the z i ). I am wondering how to pass the sign operator through the expectation.

Question

I am trying to evaluate       E    [  sign  ⁡  ⟨  v  ,  z  ⟩  ] for   v  ∈            R        n   fixed and       z    i    ∼  N  (  0  ,  1  )     ∀     i  ∈  [  n  ]The sign part is what is confusing me. Clearly,      E    [  sign  ⁡  ⟨  v  ,  z  ⟩  ]  =      E              [        sign  ⁡      ∑          i      =      1        n        v    i        z    i              ]        =      P              [            ∑          i      =      1        n        v    i        z    i    &amp;gt;  0            ]        −      P              [            ∑          i      =      1        n        v    i        z    i    &amp;lt;  0            ]        .But I don&#039;t know how to simplify further (i.e. to get the expectation just on the       z    i  ). I am wondering how to pass the sign operator through the expectation.

Tirioliwo · Accepted Answer

Observe that z and −z have the same distribution.Consequently   sign  ⁡  ⟨  v  ,  z  ⟩ has the same distribution as   sign  ⁡  ⟨  v  ,  −  z  ⟩  =  −  sign  ⁡  ⟨  v  ,  z  ⟩, so that:      E    [  sign  ⁡  ⟨  v  ,  z  ⟩  ]  =      E    [  sign  ⁡  ⟨  v  ,  −  z  ⟩  ]  =      E    [  −  sign  ⁡  ⟨  v  ,  z  ⟩  ]  =  −      E    [  sign  ⁡  ⟨  v  ,  z  ⟩  ]This implies that:      E    [  sign  ⁡  ⟨  v  ,  z  ⟩  ]  =  0

To evaluate bbbE[sign (:v,z:)] for v in bbbR^n fixed and z_i ~ N(0,1) AA i in[n]

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