How to derive the expectation of $\max(X,0)$ given the mgf function of $X$?

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I got a moment-generating function for an r.v. $X$,

$$M_X(t)=\frac{i\lambda\mu}{(\lambda +t)(\mu-t)},\forall t\in(-\lambda,\mu)$$ where $i$, $\lambda$, $\mu$ are non-negative constants.

I know the expectation of $X$ can be easily derived by $M_X^{'}(0)$.

However, I am more interested in the expectation of $\max(X,0)$. i.e., I want to derive $$\int_0^\infty x\cdot f_X(x)dx+\int_{-\infty}^0 0\cdot f_X(x)dx=\int_0^\infty x\cdot f_X(x)dx$$

based on the known $M_X(t)$.

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Let $F_1$ be the CDF of Exp$(\lambda)$ with characteristic function $ \frac{\lambda}{ \lambda - it} $ and $F_2$ the CDF of $-$Exp$(\mu)$ with characteristic function $ \frac{\mu}{ \mu + it} $.

Then the characteristic function of $\frac{\lambda}{\lambda + \mu}F_1 + \frac{\mu}{\lambda + \mu}F_2$ would be the given expression (see [1, Lemma 3.3.9]).

Now you can simply calculate the expected value of the positive part.

[1] Probability: Theory and Examples. Rick Durrett. Fifth Edition. https://services.math.duke.edu/~rtd/PTE/pte.html