CharacteristicFunctionOfXWhenWeHaveTheCharacteristicFunctionOfLogX - crowlogic/arb4j GitHub Wiki

Given the characteristic function of $Y = \ln(X)$, denoted as $\Phi_Y(t)$, we want to find the characteristic function of $X$, denoted as $\Phi_X(t)$.

First, we express $\Phi_X(t)$:

$$ \Phi_X(t) = \mathbb{E}[e^{itX}]. $$

Since $Y = \ln(X)$, we have $X = e^Y$. Substituting $X$:

$$ \Phi_X(t) = \mathbb{E}[e^{it e^Y}]. $$

Now, we use the relationship between the moment-generating function (MGF) of a random variable $W$, denoted as $M_W(t)$, and its characteristic function, denoted as $\Phi_W(t)$. The MGF of $W$ is defined as $M_W(t) = \mathbb{E}[e^{tW}]$, and the characteristic function of $W$ is $\Phi_W(t) = \mathbb{E}[e^{itW}]$. The relationship between the MGF and the characteristic function is that the MGF is the characteristic function with $t$ replaced by $-it$:

$$ M_Y(-it) = \mathbb{E}[e^{-itY}] = \Phi_Y(-t). $$

Now, we can write $\Phi_X(t)$ in terms of the MGF of $Y$:

$$ \Phi_X(t) = \mathbb{E}[e^{it e^Y}] = M_Y(-it). $$

Thus, to find $\Phi_X(t)$, we just need to replace $t$ by $-it$ in the characteristic function of $Y$:

$$ \Phi_X(t) = \Phi_Y(-it). $$

In summary, given the characteristic function of $Y = \ln(X)$, you can find the characteristic function of $X$ by replacing $t$ with $-it$ in the characteristic function of $Y$:

$$ \Phi_X(t) = \Phi_Y(-it). $$