$X_1,..,X_n$ are observations of a population with density $f(x)=\frac{1}{2}\left\{\begin{matrix} \lambda e^{\lambda x} \;\,\text{ if } x<0\\ \lambda e^{-\lambda x} \text{ if } x \geq 0 \end{matrix}\right.$
where parameter $\lambda$ is unknown. Determine a moment of moment (estimation) for $\lambda$.
Edit: I tried to answer my question with help of the given comments, see it below
I don't think they ask question like this in exam but I find it on internet and want know how example can be done correct?
How I understand is you need to calculate the expected value of the density function: $$E(X) = \int_{-\infty}^{\infty} x \cdot f(x) dx = \int_{-\infty}^{0} \frac{1}{2} x \cdot \lambda e^{\lambda x}dx + \int_{0}^{\infty} \frac{1}{2} x \cdot \lambda e^{-\lambda x}dx \\ = \left[\frac{(\lambda x-1)e^{\lambda x}}{2\lambda}\right]_{-\infty}^{0} + \left[-\frac{(\lambda x+1)e^{-\lambda x}}{2\lambda}\right]_{0}^{\infty} = \\ = \left[\frac{(\lambda \cdot 0-1)e^{\lambda \cdot 0}}{2\lambda} - \left(\frac{(\lambda \cdot (-\infty)-1)e^{\lambda \cdot (-\infty)}}{2\lambda}\right)\right]+ \\ +\left[-\frac{(\lambda \cdot \infty+1)e^{-\lambda \cdot \infty}}{2\lambda} - \left(-\frac{(\lambda \cdot 0+1)e^{-\lambda \cdot 0}}{2\lambda}\right)\right] = \\ = \left[-\frac{1}{2\lambda}-\left(\frac{-\infty \cdot 0}{2\lambda}\right)\right]+ \left[-\frac{0}{2\lambda}-\left(-\frac{1}{2\lambda}\right)\right] = \\ = \left[-\frac{1}{2\lambda}-0\right] + \left[0+\frac{1}{2\lambda}\right] = -\frac{1}{2\lambda}+\frac{1}{2\lambda}=0$$
Very frustrating after this long way all I get is zero..? :(
Is this complete wrong way of doing it? Because I don't think I do the calculation wrong at least.
So I have tried to solve this again with your help (comments):
We already calculated the first moment which is $E(X) = 0$
The second moment I solve on paper with long calculation, just assume it's correct: $E(X^2) = \frac{2}{\lambda^2}$
Thus the sample first moment is $\bar{x}= \frac{1}{n}\sum_{i=1}^{n}X_i$ and the sample second moment is $\bar{x''}= \frac{1}{n} \sum_{i=1}^{n}X_i^2$
$$E(X) = \bar{x} \Leftrightarrow 0 = \bar{x}$$
$$E(X^2) = \bar{x''} \Leftrightarrow \frac{2}{\lambda^2}= \bar{x''} \Leftrightarrow \frac{\lambda^2}{2} = \frac{1}{\bar{x''}} \Leftrightarrow \lambda^2= \frac{2}{\bar{x''}} \Leftrightarrow \lambda = ^{+}_{-}\sqrt{\frac{2}{\bar{x''}}}$$
Therefore we have method of moment estimation $$\lambda_1= \sqrt{\frac{2}{\bar{x''}}} \text{ and } \lambda_2 = -\sqrt{\frac{2}{\bar{x''}}}$$
I hope this is alright now?