Mean and Variance of Laplace Distribution
📂Probability Distribution Mean and Variance of Laplace Distribution X ∼ X \sim X ∼ When Laplace ( μ , b ) \operatorname{Laplace}(\mu, b) Laplace ( μ , b ) , the mean and variance of X X X are as follows.
E ( X ) = μ
E(X) = \mu
E ( X ) = μ
Var ( X ) = 2 b 2
\Var(X) = 2b^{2}
Var ( X ) = 2 b 2
Proof Direct Calculation By the definition of expectation and integration by parts ,
E ( X ) = ∫ − ∞ ∞ x 1 2 b e − ∣ x − μ ∣ / b d x = ∫ − ∞ μ x 2 b e ( x − μ ) / b d x + ∫ μ ∞ x 2 b e − ( x − μ ) / b d x = ( [ x 2 b ( b e ( x − μ ) / b ) ] − ∞ μ − ∫ − ∞ μ 1 2 e ( x − μ ) / b d x ) + ( [ x 2 b ( − b e − ( x − μ ) / b ) ] μ ∞ + ∫ μ ∞ 1 2 e − ( x − μ ) / b d x ) = ( μ 2 − [ b 2 e ( x − μ ) / b ] − ∞ μ ) + ( μ 2 + [ − b 2 e − ( x − μ ) / b ] μ ∞ ) = ( μ 2 − b 2 ) + ( μ 2 + b 2 ) = μ
\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*}
E ( X ) = − ∞ ∫ ∞ x 2 b 1 e − ∣ x − μ ∣/ b d x = − ∞ ∫ μ 2 b x e ( x − μ ) / b d x + μ ∫ ∞ 2 b x e − ( x − μ ) / b d x = [ 2 b x ( b e ( x − μ ) / b ) ] − ∞ μ − − ∞ ∫ μ 2 1 e ( x − μ ) / b d x + [ 2 b x ( − b e − ( x − μ ) / b ) ] μ ∞ + μ ∫ ∞ 2 1 e − ( x − μ ) / b d x = ( 2 μ − [ 2 b e ( x − μ ) / b ] − ∞ μ ) + ( 2 μ + [ − 2 b e − ( x − μ ) / b ] μ ∞ ) = ( 2 μ − 2 b ) + ( 2 μ + 2 b ) = μ
To obtain the variance , let’s calculate E ( X 2 ) E(X^{2}) E ( X 2 ) .
E ( X 2 ) = ∫ − ∞ ∞ x 2 1 2 b e − ∣ x − μ ∣ / b d x = ∫ − ∞ μ x 2 2 b e ( x − μ ) / b d x + ∫ μ ∞ x 2 2 b e − ( x − μ ) / b d x = ( [ x 2 2 ( e ( x − μ ) / b ) ] − ∞ μ − ∫ − ∞ μ x e ( x − μ ) / b d x ) + ( [ − x 2 2 e − ( x − μ ) / b ] μ ∞ + ∫ μ ∞ x e − ( x − μ ) / b d x ) = [ μ 2 2 − ( μ b − b 2 ) ] + [ μ 2 2 + ( μ b − b 2 ) ] = μ 2 + 2 b 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*}
E ( X 2 ) = − ∞ ∫ ∞ x 2 2 b 1 e − ∣ x − μ ∣/ b d x = − ∞ ∫ μ 2 b x 2 e ( x − μ ) / b d x + μ ∫ ∞ 2 b x 2 e − ( x − μ ) / b d x = [ 2 x 2 ( e ( x − μ ) / b ) ] − ∞ μ − − ∞ ∫ μ x e ( x − μ ) / b d x + [ − 2 x 2 e − ( x − μ ) / b ] μ ∞ + μ ∫ ∞ x e − ( x − μ ) / b d x = [ 2 μ 2 − ( μ b − b 2 ) ] + [ 2 μ 2 + ( μ b − b 2 ) ] = μ 2 + 2 b 2
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} Var ( X ) = E ( X 2 ) − E ( X ) 2 ,
Var ( X ) = E ( X 2 ) − E ( X ) 2 = μ 2 + 2 b 2 − μ 2 = 2 b 2
\Var(X) = E(X^{2}) - E(X)^{2} = \mu^{2} + 2b^{2} - \mu^{2} = 2b^{2}
Var ( X ) = E ( X 2 ) − E ( X ) 2 = μ 2 + 2 b 2 − μ 2 = 2 b 2
■
From the Moment Generating Function The moment generating function of the Laplace distribution is as follows.
m ( t ) = 1 1 − b 2 t 2 e μ t for ∣ t ∣ < 1 b
m(t) = \dfrac{1}{1 - b^{2}t^{2}} e^{\mu t} \qquad \text{for } |t| < \dfrac{1}{b}
m ( t ) = 1 − b 2 t 2 1 e μ t for ∣ t ∣ < b 1
Since the expectation is m ′ ( 0 ) m^{\prime}(0) m ′ ( 0 ) , by differentiating, we get
m ′ ( t ) = d d t ( 1 1 − b 2 t 2 ) e μ t + 1 1 − b 2 t 2 d d t e μ t = d d ( 1 − b 2 t 2 ) ( 1 1 − b 2 t 2 ) d ( 1 − b 2 t 2 ) d t e μ t + 1 1 − b 2 t 2 μ e μ t = 2 b 2 t ( 1 − b 2 t 2 ) 2 e μ t + μ 1 − b 2 t 2 e μ t
\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*}
m ′ ( t ) = d t d ( 1 − b 2 t 2 1 ) e μ t + 1 − b 2 t 2 1 d t d e μ t = d ( 1 − b 2 t 2 ) d ( 1 − b 2 t 2 1 ) d t d ( 1 − b 2 t 2 ) e μ t + 1 − b 2 t 2 1 μ e μ t = ( 1 − b 2 t 2 ) 2 2 b 2 t e μ t + 1 − b 2 t 2 μ e μ t
Thus, we obtain the following.
E ( X ) = m ′ ( 0 ) = μ
E(X) = m^{\prime}(0) = \mu
E ( X ) = m ′ ( 0 ) = μ
To find the variance, differentiate once more,
m ′ ′ ( t ) = d d t ( 2 b 2 t ( 1 − b 2 t 2 ) 2 e μ t + μ 1 − b 2 t 2 e μ t )
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)
m ′′ ( t ) = d t d ( ( 1 − b 2 t 2 ) 2 2 b 2 t e μ t + 1 − b 2 t 2 μ e μ t )
The differentiation of the second term can be directly obtained from the result above.
m ′ ′ ( t ) = d d t ( 2 b 2 t ( 1 − b 2 t 2 ) 2 e μ t ) + 2 μ b 2 t ( 1 − b 2 t 2 ) 2 e μ t + μ 2 1 − b 2 t 2 e μ t
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}
m ′′ ( t ) = d t d ( ( 1 − b 2 t 2 ) 2 2 b 2 t e μ t ) + ( 1 − b 2 t 2 ) 2 2 μ b 2 t e μ t + 1 − b 2 t 2 μ 2 e μ t
Differentiating the first term, it can be seen that when substituting 2 b 2 ( 1 − b 2 t 2 ) 2 e μ t \dfrac{2b^{2}}{(1-b^{2}t^{2})^{2}}e^{\mu t} ( 1 − b 2 t 2 ) 2 2 b 2 e μ t and t = 0 t=0 t = 0 , terms forming 0 0 0 appear. Therefore,
m ′ ′ ( 0 ) = 2 b 2 ( 1 − b 2 t 2 ) 2 e μ t + 2 μ b 2 t ( 1 − b 2 t 2 ) 2 e μ t + μ 2 1 − b 2 t 2 e μ t ∣ t = 0 = 2 b 2 + μ 2
\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*}
m ′′ ( 0 ) = ( 1 − b 2 t 2 ) 2 2 b 2 e μ t + ( 1 − b 2 t 2 ) 2 2 μ b 2 t e μ t + 1 − b 2 t 2 μ 2 e μ t t = 0 = 2 b 2 + μ 2
Thus, the variance is as follows.
Var ( X ) = E ( X 2 ) − ( E ( X ) ) 2 = m ′ ′ ( 0 ) − ( m ′ ( 0 ) ) 2 = 2 b 2
\Var(X) = E(X^{2}) - (E(X))^{2} = m^{\prime\prime}(0) - (m^{\prime}(0))^{2} = 2b^{2}
Var ( X ) = E ( X 2 ) − ( E ( X ) ) 2 = m ′′ ( 0 ) − ( m ′ ( 0 ) ) 2 = 2 b 2
■