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4.5: Other Continuous Distributions

  • Page ID
    4380
  • In this section, we introduce two families of continuous probability distributions that are commonly used.

    Exponential Distribution

    Definition \(\PageIndex{1}\)

    A random variable \(X\) has an exponential distribution with parameter \(\lambda>0\), write \(X\sim\text{exponential}(\lambda)\), if \(X\) has pdf given by
    $$f(x) = \left\{\begin{array}{l l}
    \lambda e^{-\lambda x}, & \text{for}\ x\geq 0, \\
    0 & \text{otherwise.}
    \end{array}\right.\notag$$

    Example \(\PageIndex{1}\)

    A typical application of the exponential distribution is to model waiting times or lifetimes. For example, each of the following gives an application of the exponential distribution.

    • \(X=\) lifetime of a radioactive particle
    • \(X=\) how long you have to wait for an accident to occur at a given intersection
    • \(X=\) length of interval between consecutive occurrences of Poisson distributed events

    The parameter \(\lambda\) is referred to as the rate parameter, it represents how quickly events occur.  For example, in the first case above where \(X\) denotes the lifetime of a radioactive particle, \(\lambda\) would give the rate at which such particles decay.

    Properties of the Exponential Distribution

    If \(X\sim\text{exponential}(\lambda)\), then the following hold.

    1. The cdf of \(X\) is given by
      $$F(x) = \left\{\begin{array}{l l}
      0 & \text{for}\ x< 0, \\
      1- e^{-\lambda x}, & \text{for}\ x\geq 0. \\
      \end{array}\right.\notag$$
    2. For any \(0 < p < 1\), the \((100p)^{\text{th}}\) percentile is \(\displaystyle{\pi_p = \frac{-\ln(p)}{\lambda}}\).
    3. The mean of \(X\) is \(\displaystyle{\text{E}[X] = \frac{1}{\lambda}}\).
    4. The variance of \(X\) is \(\displaystyle{\text{Var}(X) = \frac{1}{\lambda^2}}\).
    5. \(X\) satisfies the Memoryless Property, i.e., \(P(X>t+s\ |\ X>s) = P(X>t)\), for any \(t,s \geq0\).
    Partial Proof

    We prove Properties #1 & #3, the others are left as an exercise.

    For the first property, we consider two cases based on the value of \(x\). First, if \(x<0\), then the pdf is constant and equal to 0, which gives the following for the cdf:
    $$F(x) = \int^{x}_{-\infty} f(t) dt = \int^x_{-\infty} 0 dt = 0 \notag$$
    Second, if \(x\geq0\), then the pdf is \(\lambda e^{-\lambda x}\), and the cdf is given by $$F(x) = \int^x_{-\infty} f(t) dt = \int^x_0 \lambda e^{-\lambda t} dt = -e^{-\lambda t}\Big|^x_0 = -e^{-\lambda x} - (-e^0) = 1-e^{-\lambda x}. \notag$$

    For the third property, we Definition 4.2.1 to calculate the expected value of a continuous random variable:
    $$\text{E}[X] = \int^{\infty}_{-\infty} x\cdot f(x) dx = \int^{\infty}_0 x\cdot \lambda e^{-\lambda x} dx = -x\cdot e^{-\lambda x}\big|^{\infty}_0 + \int^{\infty}_0 e^{-\lambda x} dx = 0 + \frac{-e^{-\lambda x}}{\lambda}\big|^{\infty}_0 = \frac{1}{\lambda}. \notag$$

    In words, the Memoryless Property of the exponential distribution states that, given that you have already waited more than \(s\) units of time (\(X>s\), the conditional probability that you will have to wait \(t\) more (\(X>t+s\)) is equal to the unconditional probability you just have to wait more than \(t\) units of time. For example, suppose you are waiting for the bus and the amount of time you have to wait is exponentially distributed. If you have already been waiting 5 minutes at the bus stop, the probability that you have to wait 4 more minutes (so more than 9 minutes total) is equal to the probability that you only had to wait more than 4 minutes once arriving at the bus stop. In calculating the conditional probability, the exponential distribution "forgets" about the condition or the time already spent waiting and you can just calculate the unconditional probability that you have to wait longer.

    Gamma Distribution

    Definition \(\PageIndex{2}\)

    A random variable \(X\) has a gamma distribution with parameters \(\alpha, \lambda>0\), write \(X\sim\text{gamma}(\alpha, \lambda)\), if \(X\) has pdf given by
    $$f(x) = \left\{\begin{array}{l l}
    \displaystyle{\frac{\lambda^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\lambda x}}, & \text{for}\ x\geq 0, \\
    0 & \text{otherwise,}
    \end{array}\right. \notag$$
    where \(\Gamma(\alpha)\) is a function (referred to as the gamma function) given by the following integral:
    $$\Gamma(\alpha) = \int^{\infty}_0 t^{\alpha-1}e^{-t}dt. \notag$$

    Note that the gamma function, \(\Gamma(\alpha)\), ensures that the gamma pdf is valid, i.e., that it integrates to \(1\), which you are asked to show in the following exercise.

    Exercise \(\PageIndex{1}\)

    Show: \(\displaystyle{\int^{\infty}_0 \frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x} dx = 1}\)

    Answer

    In the integral, we can make the substitution: \(u = \lambda x \rightarrow du = \lambda dx\). Therefore, we have
    $$\int^{\infty}_0 \frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x} dx = \int^{\infty}_0 \frac{\lambda \lambda^{\alpha-1}}{\Gamma(\alpha)} x^{\alpha-1}e^{-\lambda x} dx = \frac{1}{\Gamma(\alpha)}\int^{\infty}_0 u^{\alpha-1}e^{-u} du = \frac{1}{\Gamma(\alpha)}\Gamma(\alpha) = 1. \notag$$

    Notes about the Gamma Distribution:

    • If \(\alpha = 1\), then the corresponding gamma distribution is given by the exponential distribution, i.e., \(\text{gamma}(1,\lambda) = \text{exponential}(\lambda)\). This is left as an exercise for the reader.
    • The parameter \(\alpha\) is referred to as the shape parameter, and \(\lambda\) is the rate parameter (or scale parameter).
    • A closed form does not exist for the cdf of a gamma distribution, computer software must be used to calculate gamma probabilities.

    Example \(\PageIndex{2}\)

    A typical application of the gamma distribution is to model the time it takes for a given number of events to occur. For example, each of the following gives an application of the gamma distribution.

    • \(X=\) lifetime of 5 radioactive particles
    • \(X=\) how long you have to wait for 3 accidents to occur at a given intersection

    In these examples, the parameter \(\lambda\) represents the rate at which the event occurs, and the parameter \(\alpha\) is the number of events desired. So, in the first example, \(\alpha=5\) and \(\lambda\) represents the rate at which particles decay.