# Expected Value Evaluation

1. Sep 13, 2009

### S_David

Hello,

I have the following expression:

$$G^2=\frac{\mathcal{E}_2}{\mathcal{E}_1\,\alpha+\mathcal{N}}$$

where $$\alpha$$ is an exponentially distributed random variable, and all other variables are constants. The authors said that, this expected value can be written as:

$$G^2=\int_0^{\infty}\frac{\mathcal{E}_2}{\mathcal{N}(\gamma+1)}\frac{1}{\overline{\gamma}}\text{e}^{-\gamma/\overline{\gamma}}\,d\gamma$$

Is this right, or there are typos?

Regards

2. Sep 13, 2009

### mathman

It looks like something is missing. You have a constant (E1) in the first expression which is not in the second, while you have a constant in the second (gamma bar) which is not in the first.

3. Sep 15, 2009

Here

$$G^2=\int_0^{\infty}\frac{\mathcal{E}_2}{\mathcal{N }(\gamma+1)}\frac{1}{\overline{\gamma}}\text{e}^{-\gamma/\overline{\gamma}}\,d\gamma$$

* Since the original expression was named $$G^2$$, this should be something like $$E[G^2]$$ or $$\mu_{G^2}$$

* It appears that the variable of integration is $$\gamma$$ and $$\overline \gamma$$ refers to the mean of the exponential distribution. With the form of the denominator in the integral, it does appear that either the constant from the initial expression is missing, or that it was incorrectly typed : could it be that the expression was supposed to be

$$G^2 = \frac{\mathcal{E}_2}{\mathcal{N}\alpha + \mathcal{N}} = \frac{\mathcal{E}_2}{\mathcal{N}\left( \alpha + 1 \right)} \, \text{\huge{?}}$$

If so, that explains the form of the integral.

4. Sep 15, 2009

### S_David

I am sorry, I was wrong, it says that:

$$G^2=E_{\alpha}\left[\frac{\mathcal{E}_2}{\mathcal{E}_1\,\alpha+\mathcal{N}}\right]$$

where $$E_X[.]$$ is the expectation operator. Anyway, the authors has the aforementioned integral evaluated as:

$$G^2=\frac{\mathcal{E}_2}{\mathcal{E}_1\,\Omega_1}\,\text{e}^{1/\overline{\gamma}}\,E_1\left(\frac{1}{\overline{\gamma}}\right)$$

where $$\overline{\gamma}=\frac{\Omega_1\,\mathcal{E}_1}{\mathcal{N}}$$, and $$E_1(.)$$ is the exponential integral. The final result is consistent with the integration, so, this removes any doubt about typos, doesn't it?

5. Sep 15, 2009

So, your post was the typo? doesn't matter, as long as the confusion is cleared up for you. glad you got to the bottom of it.

6. Sep 15, 2009

### S_David

No, I posted what they wrote, except the expectation operator that I forgot. What I meant is that it is unlikely that the authors continue on typos. So, the matter is still stucked for me. I don't know what is the missing part in their solution.