📂Probability Distribution

Mean and Variance of Laplace Distribution

Formula

$X \sim$ When $\operatorname{Laplace}(\mu, b)$, the mean and variance of $X$ are as follows.

$$E(X) = \mu$$ $$\Var(X) = 2b^{2}$$

Proof

Direct Calculation

By the definition of expectation and integration by parts,

$$\begin{align*} E(X) &= \int\limits_{-\infty}^{\infty} x \dfrac{1}{2b} e^{-|x - \mu|/b} dx \\ &= \int\limits_{-\infty}^{\mu} \dfrac{x}{2b} e^{(x - \mu)/b} dx + \int\limits_{\mu}^{\infty} \dfrac{x}{2b} e^{-(x - \mu)/b} dx \\ &= \left( \left[\dfrac{x}{2b}\left(b e^{(x - \mu)/b}\right)\right]_{-\infty}^{\mu} - \int\limits_{-\infty}^{\mu} \dfrac{1}{2} e^{(x - \mu)/b} dx \right) \\ &\qquad+ \left( \left[\dfrac{x}{2b}\left(-b e^{-(x - \mu)/b}\right)\right]_{\mu}^{\infty} + \int\limits_{\mu}^{\infty} \dfrac{1}{2} e^{-(x - \mu)/b} dx \right) \\ &= \left( \dfrac{\mu}{2} - \left[ \dfrac{b}{2} e^{(x - \mu)/b} \right]_{-\infty}^{\mu} \right) + \left( \dfrac{\mu}{2} + \left[ -\dfrac{b}{2} e^{-(x-\mu)/b} \right]_{\mu}^{\infty} \right) \\ &= \left( \dfrac{\mu}{2} - \dfrac{b}{2} \right) + \left( \dfrac{\mu}{2} + \dfrac{b}{2} \right) \\ &= \mu \end{align*}$$

To obtain the variance, let’s calculate $E(X^{2})$.

$$\begin{align*} E(X^{2}) &= \int\limits_{-\infty}^{\infty} x^{2} \dfrac{1}{2b} e^{-|x - \mu|/b} dx \\ &= \int\limits_{-\infty}^{\mu} \dfrac{x^{2}}{2b} e^{(x - \mu)/b} dx + \int\limits_{\mu}^{\infty} \dfrac{x^{2}}{2b} e^{-(x - \mu)/b} dx \\ &= \left( \left[\dfrac{x^{2}}{2}\left( e^{(x - \mu)/b}\right)\right]_{-\infty}^{\mu} - \int\limits_{-\infty}^{\mu} x e^{(x - \mu)/b} dx \right) \\ &\qquad+ \left( \left[ - \dfrac{x^{2}}{2} e^{-(x - \mu)/b}\right]_{\mu}^{\infty} + \int\limits_{\mu}^{\infty} x e^{-(x - \mu)/b} dx \right) \\ &= \left[ \dfrac{\mu^{2}}{2} - (\mu b - b^{2}) \right] + \left[ \dfrac{\mu^{2}}{2} + (\mu b - b^{2}) \right] \\ &= \mu^{2} + 2b^{2} \end{align*}$$

The third equality is due to integration by parts, and the fourth equality uses the result of the integration calculated for the expectation. Since variance is $\Var(X) = E(X^{2}) - E(X)^{2}$,

$$\Var(X) = E(X^{2}) - E(X)^{2} = \mu^{2} + 2b^{2} - \mu^{2} = 2b^{2}$$

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From the Moment Generating Function

The moment generating function of the Laplace distribution is as follows.

$$m(t) = \dfrac{1}{1 - b^{2}t^{2}} e^{\mu t} \qquad \text{for } |t| < \dfrac{1}{b}$$

Since the expectation is $m^{\prime}(0)$, by differentiating, we get

$$\begin{align*} m^{\prime}(t) &= \dfrac{d}{dt} \left( \dfrac{1}{1 - b^{2}t^{2}} \right) e^{\mu t} + \dfrac{1}{1 - b^{2}t^{2}} \dfrac{d}{dt} e^{\mu t} \\ &= \dfrac{d}{d(1-b^{2}t^{2})} \left( \dfrac{1}{1 - b^{2}t^{2}} \right) \dfrac{d(1-b^{2}t^{2})}{dt} e^{\mu t} + \dfrac{1}{1 - b^{2}t^{2}} \mu e^{\mu t} \\ &= \dfrac{2b^{2}t}{(1 - b^{2}t^{2})^{2}} e^{\mu t} + \dfrac{\mu}{1 - b^{2}t^{2}} e^{\mu t} \end{align*}$$

Thus, we obtain the following.

$$E(X) = m^{\prime}(0) = \mu$$

To find the variance, differentiate once more,

$$m^{\prime\prime}(t) = \dfrac{d}{dt} \left( \dfrac{2b^{2}t}{(1 - b^{2}t^{2})^{2}} e^{\mu t} + \dfrac{\mu}{1 - b^{2}t^{2}} e^{\mu t} \right)$$

The differentiation of the second term can be directly obtained from the result above.

$$m^{\prime\prime}(t) = \dfrac{d}{dt} \left( \dfrac{2b^{2}t}{(1 - b^{2}t^{2})^{2}} e^{\mu t} \right) + \dfrac{2\mu b^{2}t}{(1 - b^{2}t^{2})^{2}} e^{\mu t} + \dfrac{\mu^{2}}{1 - b^{2}t^{2}} e^{\mu t}$$

Differentiating the first term, it can be seen that when substituting $\dfrac{2b^{2}}{(1-b^{2}t^{2})^{2}}e^{\mu t}$ and $t=0$, terms forming $0$ appear. Therefore,

$$\begin{align*} m^{\prime\prime}(0) &= \left. \dfrac{2b^{2}}{(1-b^{2}t^{2})^{2}}e^{\mu t} + \dfrac{2\mu b^{2}t}{(1 - b^{2}t^{2})^{2}} e^{\mu t} + \dfrac{\mu^{2}}{1 - b^{2}t^{2}} e^{\mu t} \right|_{t=0} \\ &= 2b^{2} + \mu^{2} \\ \end{align*}$$

Thus, the variance is as follows.

$$\Var(X) = E(X^{2}) - (E(X))^{2} = m^{\prime\prime}(0) - (m^{\prime}(0))^{2} = 2b^{2}$$

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