Can the Expected Value be Written as an Integral?

AI Thread Summary
The discussion centers around the expression for G^2 and whether it can be accurately represented as an integral. Initially, there was confusion regarding the constants present in the original expression compared to the integral form. It was clarified that the correct representation includes the expectation operator, leading to the expression G^2=E[...]. The authors provided an evaluated integral that aligns with the expected value, suggesting no typos in their work. However, one participant remains uncertain about a missing component in the authors' solution.
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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
 
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S_David said:
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

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.
 
Here

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

* 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

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

If so, that explains the form of the integral.
 
statdad said:
Here

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

* 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

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

If so, that explains the form of the integral.

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?
 
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.
 
statdad said:
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.

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.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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