Sentence examples for i e m from inspiring English sources

Similarly, the mutual information between the source S m and the eavesdropper (i.e., the untrusted relay R) on the n th subcarrier, denoted by I e, m n, can be written as I e, m n = I r, m n = log 2 1 + p s h a, m n 2 σ 2 (2).

It is easy to show that the pdf of z i = log(f i ) is f z i = ( m i ) m i Γ ( m i ) e − m i e z i e m i z i. Denote its mean and variance by E[z i ] and Var[z i ], respectively.

So, a plug-in estimator of the nonparametric component g, based on β ^ n, is given by ĝ n ( t ) ≜ ĝ n ( t, β ^ n ) = ∑ i = 1 n ( Y i - x i β ^ n ) ∫ A i E m ( t, s ) d s. (1.5).

If the high SNR scenario is assumed, according to [1], the secrecy rate of the m th pair on the n th subcarrier can be expressed as RS m n = I d, m n − I e, m n + = log 2 h d, m n 2 h a, m n 2 + p m n h b, m n 2 p s h a, m n 2 + p m n h b, m n 2 + (3).

Because α ∈ ℝ + *, a given emission intensity I e m 634 can correspond to any [PpIX].

Considering the presence of two states for PpIX in vivo, [PpIX] cannot be retrieved only from the emission intensity (I e m 634 ) at 634 nm.

where f(h1,…,h M ) is the pdf for the channel vector,given by f ( h 1, …, h M ) = ∏ i = 1 M ( m i λ i ) m i h i m i − 1 Γ ( m i ) e − m i λ i h i for Nakagami fading (due to independence amongst the parallel channels).

For the purpose of analysis, we will assume that each channel h i is gamma distributed (Nakagami fading) with mean 1 λ i, which probability density function (pdf) is given by f ( h i ) = ( m i λ i ) m i h i m i − 1 Γ ( m i ) e − m i λ i h i, h i > 0, (5).

Assuming that either Q1 or Q2 holds, this expression is zero if and only if we are in the intersection of the two model classes: I (E 2 ; M 1 | M 2 ) = 0, I (E 1 ; M 2 | M 1 ) = 0, I (E 2 ; M 1 | M 2, E 1 ) = 0, I (E 1 ; M 2 | M 1, E 2 ) = 0.

We can expand the left hand side of Eq. 34 by using the definitions M ⊺ v i = ∑ l ∈ 1 M v i ( l ) and v i ⊺ ζ k = v i ( l + ) − v i ( l − ), which yield ∑ i = 2 n ∑ j ≠ i E [ M ⊺ v i v i ⊺ ζ k ] E [ M ⊺ v j v j ⊺ ζ k ] (38).

We construct g i (x) as g i ( x ) = lcm { M i e 1 ( x ), M i e 2 ( x ), ⋯, M i e n i − k i ( x ) }, Open image in new window.