Log-normal shadowing and mean power

Question

Let's assume I'm receiving a signal with power measured in dBm shadowed according to the normal distribution, i.e. $P_{\text{dBm}} \sim N(0,1) $.

If there is also a normally distributed noise power $N_{\text{dBm}}$ of mean $0$ dBm present, the signal-to-noise ratio is \begin{equation} \label{hur} \text{SNR}_{\text{dB}} = \mathbb{E}[P_{\text{dBm}} - N_{\text{dBm}}] = \mathbb{E}[P_{\text{dBm}}] -\mathbb{E}[N_{\text{dBm}}]= 0. \hspace{1cm} (1) \end{equation}

In the other hand, both noise and signal power measured in mW follows a log-normal distribution (base 10) $P_{\text{mW}} , N_{\text{mW}} \sim \text{Lognormal}(0,1/10), $ and their sum follow $P_{\text{mW}} + N_{\text{mW}} \sim \text{Lognormal}(0,1/50)$.

We can derive the linear SNR as $$ \text{SNR} = \mathbb{E}\left[\frac{10^{P_{\text{dBm}}/10}}{10^{N_{\text{dBm}/10}}} \right] = \mathbb{E}[10^{\frac{P_{\text{dBm}}-N_{\text{dBm}}}{10}}] = e^{\frac{ln(10)^2}{100}} \approx 1.05, $$ which should be equivalent to (1), but clearly it is not; $10\log(1.05) \approx 0.21 > 0$.

I understand that this kind of relates to the fact that in this case $\mathbb{E}[10^{P_{\text{dBm}}/10 }] \neq 10^{\mathbb{E}[P_{\text{dBm}}]/10 },$ but it does not make sense to me that the channel is better in the latter case where the expected power is treated through the log-normal distribution. Channel quality seems to depend on the unit we measure the signal power.

So my question is; what I am missing here? How is this to be interpreted in the real world?

Answer 1

I guess that the confusion was exactly that $$ \mathbb{E}[10^{P_\text{dBm}/10}] \neq 10^{\mathbb{E}[P_\text{dBm}]/10}. $$

Even though the expected dBm under normally distributed shadowing would be $0$, we just have to accept that the expected mW is > 1. The higher mWs are given more weight than the lower values.

We can't measure the mean of dBm, and then conclude that mean measured in mW would be 1.