| # | x | μ | b | F(x) | 1 − F(x) | Actions |
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Formula Used
The Laplace distribution with location \( \mu \) and scale \( b>0 \) has pdf \( f(x) = \dfrac{1}{2b}\exp\!\big(-\tfrac{|x-\mu|}{b}\big) \).
Its cumulative distribution function (CDF) is piecewise:
For \(x<\mu\): \( F(x)=\tfrac{1}{2}\exp\!\big(\tfrac{x-\mu}{b}\big) \).
For \(x\ge \mu\): \( F(x)=1-\tfrac{1}{2}\exp\!\big(-\tfrac{x-\mu}{b}\big) \).
The complementary probability is \( \bar{F}(x)=1-F(x) \). Mean is \( \mu \); variance is \( 2b^2 \).
How to Find CDF for Laplace Distribution
Start from the pdf \( f(x)=\dfrac{1}{2b}\exp\!\big(-\tfrac{|x-\mu|}{b}\big) \). The CDF is \(F(x)=\int_{-\infty}^{x} f(t)\,dt\). Because of the absolute value, integrate in two cases.
For \(x<\mu\): \; F(x)=\displaystyle\int_{-\infty}^{x}\frac{1}{2b}\exp\!\Big(\frac{t-\mu}{b}\Big)\,dt = \frac{1}{2}\exp\!\Big(\frac{x-\mu}{b}\Big).\)
For \(x\ge\mu\): \; F(x)=\displaystyle\int_{-\infty}^{\mu}\frac{1}{2b}\exp\!\Big(\frac{t-\mu}{b}\Big)\,dt + \int_{\mu}^{x}\frac{1}{2b}\exp\!\Big(-\frac{t-\mu}{b}\Big)\,dt = 1-\frac{1}{2}\exp\!\Big(-\frac{x-\mu}{b}\Big).\)
A compact form is \( F(x)=\tfrac{1}{2} + \tfrac{1}{2}\,\mathrm{sgn}(x-\mu)\big(1-\exp(-|x-\mu|/b)\big) \). Here \( \mathrm{sgn} \) is the sign function.
Quick numerical checks
- \(\mu=0, b=2, x=1:\; F(1)=1-\tfrac{1}{2}e^{-0.5}\approx 0.69673467.\)
- \(\mu=0, b=2, x=-3:\; F(-3)=\tfrac{1}{2}e^{-1.5}\approx 0.11156508.\)
How to Use This Calculator
- Enter the evaluation point \(x\), location \(\mu\), and scale \(b>0\).
- Choose precision for rounding. Optionally show \(1-F(x)\).
- Click Compute for single value results and a permalink.
- For datasets, add rows or import a CSV with columns: x, mu, b.
- Export your table as CSV or generate a PDF snapshot for reporting.
Example Data Table
Click Load example to insert the following rows and reproduce results on your side:
- \(\mu=0, b=2\) with \(x\in\{-3,-1,0,1,3\}\).