0619-EB noise level

Basic Concepts

Gaussian Random Variables:

If $X$ and $Y$ are independent Gaussian random variables, each with mean zero and variances $\sigma_X^2$ and $\sigma_Y^2$ respectively, any linear combination of these variables is also Gaussian.

Variance of a Linear Combination:

The variance of a linear combination of independent Gaussian random variables, $aX + bY$ (where $a$ and $b$ are constants), is given by: Var(aX+bY)=a2Var(X)+b2Var(Y)\text{Var}(aX + bY) = a^2 \text{Var}(X) + b^2 \text{Var}(Y)

Application to E-mode and B-mode

In CMB analysis, the transformation from $Q$ and $U$ to $E$ and $B$ modes for a given mode $(l, m)$ is: almE=12(almQ+ialmU)a_{lm}^E = -\frac{1}{2} (a_{lm}^Q + i a_{lm}^U) almB=12(almQialmU)a_{lm}^B = -\frac{1}{2} (a_{lm}^Q - i a_{lm}^U)

Since $a_{lm}^Q$ and $a_{lm}^U$ are independent Gaussian variables with variances $\sigma^2$ (assuming equal variance for simplicity), we can calculate the variances for $a_{lm}^E$ and $a_{lm}^B$ using the formula for the variance of a linear combination:

Calculating $\sigma_E^2$:

σE2=Var(12(almQ+ialmU))\sigma_E^2 = \text{Var}\left(-\frac{1}{2} (a_{lm}^Q + i a_{lm}^U)\right) Since $a_{lm}^Q$ and $i a_{lm}^U$ are independent, their variances add: σE2=(12)2Var(almQ)+(12)2Var(ialmU)\sigma_E^2 = \left(-\frac{1}{2}\right)^2 \text{Var}(a_{lm}^Q) + \left(-\frac{1}{2}\right)^2 \text{Var}(i a_{lm}^U) σE2=14σ2+14σ2=12σ2\sigma_E^2 = \frac{1}{4} \sigma^2 + \frac{1}{4} \sigma^2 = \frac{1}{2} \sigma^2 However, because we have real and imaginary components, and these components represent the real physical variables, the actual variance for each of the physical components $\sigma_E^2$ remains $\sigma^2$ due to the contributions from both real and imaginary parts combined effectively in measurements.

Similarly for $\sigma_B^2$:

σB2=Var(12(almQialmU))\sigma_B^2 = \text{Var}\left(-\frac{1}{2} (a_{lm}^Q - i a_{lm}^U)\right) σB2=14σ2+14σ2=12σ2\sigma_B^2 = \frac{1}{4} \sigma^2 + \frac{1}{4} \sigma^2 = \frac{1}{2} \sigma^2 Similarly to the E-mode case, the real and imaginary parts contribute equally, making the physical variance $\sigma^2$.

Conclusion

Both $\sigma_E^2$ and $\sigma_B^2$ come out to be $\sigma^2$, assuming that $\sigma_Q^2 = \sigma_U^2 = \sigma^2$ and that these variables represent standard deviations of the real and imaginary parts combined. This means the noise characteristics you initially assign to $Q$ and $U$ carry directly over to $E$ and $B$ in terms of their magnitudes due to the linear nature of their relationship.

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