# Transformed gamma distribution

The previous post opens up a discussion on generating distributions by raising an existing distribution to a power. The previous post focuses on the example of raising an exponential distribution to a power. This post focuses on the distributions generated by raising a gamma distribution to a power.

Raising to a Power

Let $X$ be a random variable. Let $\tau$ be a positive constant. The random variables $Y=X^{1/\tau}$, $Y=X^{-1}$ and $Y=X^{-1/\tau}$ are called transformed, inverse and inverse transformed, respectively.

Let $f_X(x)$, $F_X(x)$ and $S_X(x)=1-F_X(x)$ be the probability density function (PDF), the cumulative distribution function (CDF) and the survival function of the random variable $X$ (the base distribution). The following derivation seeks to express the CDFs of the “transformed” variables in terms of the base CDF $F_X(x)$.

$Y=X^{1/\tau}$ (Transformed):
\displaystyle \begin{aligned} F_Y(y)&=P(Y \le y) \\&=P(X^{1/\tau} \le y) \\&=P(X \le y^\tau) \\&=F_X(y^\tau) \end{aligned}

$Y=X^{-1}$ (Inverse):
\displaystyle \begin{aligned} F_Y(y)&=P(Y \le y) \\&=P(X^{-1} \le y) \\&=P(X \ge y^{-1}) \\&=S_X(y^{-1})=1-F_X(y^{-1}) \end{aligned}

$Y=X^{-1/\tau}$ (Inverse Transformed):
\displaystyle \begin{aligned} F_Y(y)&=P(Y \le y) \\&=P(X^{-1/\tau} \le y)\\&=P(X^{-1} \le y^\tau) \\&=P(X \ge y^{-\tau)} \\&=S_X(y^{-\tau})=1-F_X(y^{-\tau}) \end{aligned}

Gamma CDF

Unlike the exponential distribution, the CDF of the gamma distribution does not have a closed form. Suppose that $X$ is a random variable that has a gamma distribution with shape parameter $\alpha$ and scale parameter $\theta$. The following is the expression of the gamma CDF.

$\displaystyle F_X(x)=\int_0^x \frac{1}{\Gamma(\alpha)} \ \frac{1}{\theta^\alpha} \ t^{\alpha-1} \ e^{- t/\theta} \ dt \ \ \ \ \ \ \ \ \ x>0$

By a change of variable, the CDF can be expressed as the following integral.

\displaystyle \begin{aligned} F_X(x)&=\int_0^{x/\theta} \frac{1}{\Gamma(\alpha)} \ u^{\alpha-1} \ e^{- u} \ du \ \ \ \ \ \ \ \ \ x>0 \\&=\frac{1}{\Gamma(\alpha)} \ \gamma(\alpha; x/\theta) \end{aligned}

Note that $\Gamma(\cdot)$ and $\gamma(\beta;\cdot)$ are the gamma function and incomplete gamma function, respectively, defined as follows.

$\displaystyle \Gamma(\beta)=\int_0^\infty \ t^{\beta-1} \ e^{- t} \ dt$

$\displaystyle \gamma(\beta; w)=\int_0^w \ t^{\beta-1} \ e^{- t} \ dt$

The CDF $F_X(x)$ can be evaluated numerically using software. When the gamma distribution is raised to a power, the resulting CDF will be defined as a function of $F_X(x)$.

“Transformed” Gamma CDFs

The CDF of the “transformed” gamma distributions does not have a closed form. Thus the CDFs are to be defined based on an integral or the incomplete gamma function, shown in the preceding section. We still use the two-step approach – first deriving the CDF without the scale parameter and then add it at the end. Based on the two preceding sections, the following shows the CDFs of the three different cases.

Step 1. “Transformed” Gamma CDF (without scale parameter)

Transformed $Y=X^{1 / \tau}$ $\tau >0$
$F_Y(y)=F_X(y^\tau)$
$\displaystyle F_Y(y)=\int_0^{y^\tau} \frac{1}{\Gamma(\alpha)} \ u^{\alpha-1} \ e^{- u} \ du$ $y >0$
$\displaystyle F_Y(y)=\frac{1}{\Gamma(\alpha)} \ \gamma(\alpha; y^\tau)$ $y >0$
Inverse $Y=X^{1 / \tau}$ $\tau=-1$
$F_Y(y)=S_X(y^{-1})=1-F_X(y^{-1})$
$\displaystyle F_Y(y)=1-\int_0^{y^{-1}} \frac{1}{\Gamma(\alpha)} \ u^{\alpha-1} \ e^{- u} \ du$ $y >0$
$\displaystyle F_Y(y)=1-\frac{1}{\Gamma(\alpha)} \ \gamma(\alpha; y^{-1})$ $y >0$
Inverse Transformed $Y=X^{-1 / \tau}$ $\tau >0$
$F_Y(y)=S_X(y^{-\tau})=1-F_X(y^{-\tau})$
$\displaystyle F_Y(y)=1-\int_0^{y^{-\tau}} \frac{1}{\Gamma(\alpha)} \ u^{\alpha-1} \ e^{- u} \ du$ $y >0$
$\displaystyle F_Y(y)=1-\frac{1}{\Gamma(\alpha)} \ \gamma(\alpha; y^{-\tau})$ $y >0$

Step 2. “Transformed” Gamma CDF (with scale parameter)

Transformed $Y=X^{1 / \tau}$ $\tau >0$
$\displaystyle F_Y(y)=\int_0^{(y/\theta)^\tau} \frac{1}{\Gamma(\alpha)} \ u^{\alpha-1} \ e^{- u} \ du$ $y >0$
$\displaystyle F_Y(y)=\frac{1}{\Gamma(\alpha)} \ \gamma(\alpha; (y/\theta)^\tau)$ $y >0$
Inverse $Y=X^{1 / \tau}$ $\tau=-1$
$\displaystyle F_Y(y)=1-\int_0^{(y/\theta)^{-1}} \frac{1}{\Gamma(\alpha)} \ u^{\alpha-1} \ e^{- u} \ du$ $y >0$
$\displaystyle F_Y(y)=1-\frac{1}{\Gamma(\alpha)} \ \gamma(\alpha; (y/\theta)^{-1})$ $y >0$
Inverse Transformed $Y=X^{-1 / \tau}$ $\tau >0$
$\displaystyle F_Y(y)=1-\int_0^{(y/\theta)^{-\tau}} \frac{1}{\Gamma(\alpha)} \ u^{\alpha-1} \ e^{- u} \ du$ $y >0$
$\displaystyle F_Y(y)=1-\frac{1}{\Gamma(\alpha)} \ \gamma(\alpha; (y/\theta)^{-\tau})$ $y >0$

The transformed gamma distribution and the inverse transformed gamma distribution are three-parameter distributions with $\tau$ being the shape parameter, $\theta$ being the scale parameter and $\tau$ being in the power to which the base gamma distribution is raised. The inverse gamma distribution has two parameters with $\theta$ being the scale parameter and $\alpha$ being shape parameter (the same two parameters in the base gamma distribution). Computation of these CDFs would require the use of software that can evaluate incomplete gamma function.

Another Way to Work with “Transformed” Gamma

The CDFs derived in the preceding section is a two-step approach – first raising a gamma distribution with scale parameter equals to 1 to a power and then adding a scale parameter. The end result gives CDFs that are a function of the incomplete gamma function. Calculating a CDF would require using a software that has the capability of evaluating incomplete gamma function (or evaluating an equivalent integral). If the software that is used does not have the incomplete gamma function but has gamma CDF (e.g. Excel), then there is another way of generating the “transformed” gamma CDF.

Observe that the CDFs in the last section are the results of raising a base gamma distribution with shape parameter $\alpha$ and $\theta^\tau$ (transformed), shape parameter $\alpha$ and $\theta^{-1}$ (inverse) and shape parameter $\alpha$ and $\theta^{-\tau}$ (inverse transformed). Based on this observation, we can evaluate the CDFs and the moments. This section shows how to evaluate CDFs using this approach. The next section shows how to evaluate the moments.

Transformed Gamma Distribution

Given a transformed gamma random variable $Y$ with parameters $\tau$, $\alpha$ (shape) and $\theta$ (scale), know that $Y=X^{1/\tau}$ where $X$ gas a gamma distribution with parameters $\alpha$ (shape) and $\theta^\tau$ (scale). Then $F_Y(y)=F_X(y^\tau)$ such that $F_X(y^\tau)$ is evaluated using a software with the capability of evaluating gamma CDF (e.g. Excel).

Inverse Gamma Distribution

Given an inverse gamma random variable $Y$ with parameters $\tau$ and $\theta$ (scale), know that $Y=X^{-1}$ where $X$ gas a gamma distribution with parameters $\alpha$ (shape) and $\theta^{-1}$ (scale). Then $F_Y(y)=S_X(y^{-1})=1-F_X(y^{-1})$ such that $F_X(y^{-1})$ is evaluated using a software with the capability of evaluating gamma CDF (e.g. Excel).

Inverse Transformed Gamma Distribution

Given an inverse transformed gamma random variable $Y$ with parameters $\tau$, $\alpha$ (shape) and $\theta$ (scale), know that $Y=X^{-1/\tau}$ where $X$ gas a gamma distribution with parameters $\alpha$ (shape) and $\theta^{-\tau}$ (scale). Then $F_Y(y)=S_X(y^{-\tau})=1-F_X(y^{-\tau})$ such that $F_X(y^{-\tau})$ is evaluated using a software with the capability of evaluating gamma CDF (e.g. Excel).

Example 1 below uses Excel to compute the transformed gamma CDF.

“Transformed” Gamma Moments

Following the idea in the preceding section, the moments for the “transformed” gamma moments can be derived using gamma moments with the appropriate parameters (see here for the gamma moments). The following table shows the results.

“Transformed Gamma Moments

Base Gamma Parameters Moments
Transformed $\alpha$ and $\theta^\tau$ $\displaystyle E(Y^k)=E(X^{k/\tau})=\frac{\theta^k \Gamma(\alpha+k/\tau)}{\Gamma(\alpha)}$
$k>- \alpha \ \tau$
Inverse $\alpha$ and $\theta^{-1}$ $\displaystyle E(Y^k)=E(X^{-k})=\frac{\theta^k \Gamma(\alpha-k)}{\Gamma(\alpha)}$
$k<\alpha$
Inverse Transformed $\alpha$ and $\theta^{-\tau}$ $\displaystyle E(Y^k)=E(X^{-k/\tau})=\frac{\theta^k \Gamma(\alpha-k/\tau)}{\Gamma(\alpha)}$
$k<\alpha \ \tau$

Note that the moments for the transformed gamma distribution exists for all positive $k$. The moments for inverse gamma are limited (capped by the shape parameter $\alpha$). The moments for inverse transformed gamma are also limited, this time limited by both parameters. The computation for these moments is usually done by software, except when the argument of the gamma function is a positive integer.

“Transformed” Gamma PDFs

Once the CDFs are known, the PDFs are derived by taking derivative. The following table gives the three PDFs.

“Transformed” Gamma PDFs

Transformed $\displaystyle f_Y(y)=\frac{\tau}{\Gamma(\alpha)} \ \biggl( \frac{y}{\theta} \biggr)^{\tau \alpha} \ \frac{1}{y} \ \exp(-(y/\theta)^\tau)$ $y>0$
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Inverse $\displaystyle f_Y(y)=\frac{1}{\Gamma(\alpha)} \ \biggl( \frac{\theta}{y} \biggr)^{\alpha} \ \frac{1}{y} \ \exp(-\theta/y)$ $y>0$
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Inverse Transformed $\displaystyle f_Y(y)=\frac{\tau}{\Gamma(\alpha)} \ \biggl( \frac{\theta}{y} \biggr)^{\tau \alpha} \ \frac{1}{y} \ \exp(-(\theta/y)^\tau)$ $y>0$

Each PDF is obtained by taking derivative of the integral for the corresponding CDF.

Example

The post is concluded with one example demonstrating the calculation for CDF and percentiles using the gamma distribution function in Excel.

Example 1
The size of a collision claim from a large pool of auto insurance policies has a transformed gamma distribution with parameters $\tau=2$, $\alpha=2.5$ and $\theta=4$. Determine the following.

• The probability that a randomly selected claim is greater than than 6.75.
• The probability that a randomly selected claim is between 4.25 and 8.25.
• The median size of a claim from this pool of insurance policies.
• The mean and variance of a claim from this pool of insurance policies.
• The probability that a randomly selected claim is within one standard deviation of the mean claim size.
• The probability that a randomly selected claim is within two standard deviations of the mean claim size.

Since $\tau=2$, this is a transformed gamma distribution. There are two ways to get a handle on this distribution. One way is to plug in the three parameters $\tau=2$, $\alpha=2.5$ and $\theta=4$ into the transformed gamma CDF (in the table Step 2. “Transformed” Gamma CDF (with scale parameter)). This would require using a software that can evaluate the incomplete gamma function or its equivalent integrals. The other way is to know that this distribution is the result of raising the gamma distribution with shape parameter $\alpha=2.5$ and scale parameter $4^2=16$ to the power of 1/2. We take the latter approach.

In Excel, =GAMMA.DIST(A1,B1,C1,TRUE) is the function that produces the gamma CDF $F_X(x)$, assuming that the value $x$ is in cell A1, the shape parameter is in cell B1 and the scale parameter is in cell C1. When the last parameter is TRUE, it gives the CDF. If it is FALSE, it gives the PDF. For example, =GAMMA.DIST(2.5, 2, 1, TRUE) gives the value of 0.712702505.

The transformed gamma distribution in question (random variable $Y$) is the result of raising gamma $\alpha=2.5$ and scale parameter 16 (random variable $X$) to 1/2. Thus the CDF $F_Y(y)$ is obtained from evaluating the gamma CDF $F_X(y^2)$, which is =GAMMA.DIST($y^2$, 2.5, 16, TRUE). As a result, the following gives the answers for the first two bullet points.

$P(Y>6.75)=1-F_X(6.75)=1-0.663=0.337$
=GAMMA.DIST($6.75^2$, 2.5, 16, TRUE)

$P(4.25 \le Y \le 8.25)=F_X(8.25)-F_X(4.25)=0.8696-0.1876=0.6820$

The median or other percentiles of transformed gamma distribution are obtained by a trial and error approach, i.e. by plugging in values of $y$ into the gamma CDF =GAMMA.DIST($y^2$, 2.5, 16, TRUE). After performing the trial and error process, we see that $F_Y(5.9001425)=0.5$ and $F_Y(5.9101425)=0.502020715$. Thus we take the median to be 5.9. So the median size of a claim is around 5.9.

Computation of the moments of a gamma distribution requires the evaluation of the gamma function $\Gamma(\cdot)$. Excel does not have an explicit function for gamma function. Since $\Gamma(\cdot)$ is in the gamma PDF, we can derive the gamma function from the gamma PDF in Excel =GAMMA.DIST(1, a, 1, FALSE). The value of this gamma PDF is $e^{-1}/\Gamma(a)$. This the value of $\Gamma(a)$ is $e^{-1}$ divided by the value of this gamma PDF value. For example, the following Excel formulas give $\Gamma(5)$ and $\Gamma(2.5)$.

=EXP(-1)/GAMMA.DIST(1, 5, 1, FALSE) = 24

=EXP(-1)/GAMMA.DIST(1, 2.5, 1, FALSE) = 1.329340388

The moments $E(Y^k)$ is $E(X^{k/2})$. Based on the results in the section on “Transformed” gamma moments, the following gives the mean and variance.

$\displaystyle E(Y)=E(X^{1/2})=\frac{16^{1/2} \ \Gamma(2.5+1/2)}{\Gamma(2.5)}=6.018022225$

$\displaystyle E(Y^2)=E(X)=2.5 \cdot 16=40$

$Var(Y)=40-6.018022225^2=3.783408505$

$\sigma_Y=3.783408505^{0.5}=1.945098585$

The following gives the probability of a claim within one or two standard deviations of the mean.

\displaystyle \begin{aligned} P(\mu-\sigma

\displaystyle \begin{aligned} P(\mu-2 \sigma

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$\copyright$ 2017 – Dan Ma

# Transformed exponential distributions

The processes of creating distributions from existing ones are an important topic in the study of probability models. Such processes expand the tool kit in the modeling process. Two examples: new distributions can be generated by taking independent sum of old ones or by mixing distributions (the result would be called a mixture). Another way to generate distributions is through raising a distribution to a power, which is the subject of this post. Start with a random variable $X$ (the base distribution). Then raising it to a constant generates a new distribution. In this post, the base distribution is the exponential distribution. The next post discusses transforming the gamma distribution.

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Raising to a Power

Let $X$ be a random variable. Let $\tau$ be a nonzero constant. The new distribution is generated when $X$ is raised to the power of $1 / \tau$. Thus the random variable $Y=X^{1 / \tau}$ is the subject of the discussion in this post.

When $\tau >0$, the distribution for $Y=X^{1 / \tau}$ is called transformed. When $\tau=-1$, the distribution for $Y=X^{1 / \tau}$ is called inverse. When $\tau <0$ and $\tau \ne -1$, the distribution for $Y=X^{1 / \tau}$ is called inverse transformed.

If the base distribution is exponential, then raising it to $1 / \tau$ would produce a transformed exponential distribution for the case of $\tau >0$, an inverse exponential distribution for the case of $\tau=-1$ and an inverse transformed exponential distribution for the case $\tau <0$ with $\tau \ne -1$. If the base distribution is a gamma distribution, the three new distributions would be transformed gamma distribution, inverse gamma distribution and inverse transformed gamma distribution.

For the case of inverse transformed, we make the random variable $Y=X^{-1 / \tau}$ by letting $\tau >0$. The following summarizes the definition.

Name of Distribution Parameter $\tau$ Random Variable
Transformed $\tau >0$ $Y=X^{1 / \tau}$
Inverse $\tau=-1$ $Y=X^{1 / \tau}$
Inverse Transformed $\tau >0$ $Y=X^{-1 / \tau}$

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Transforming Exponential

The “transformed” distributions discussed here have two parameters, $\tau$ and $\theta$ ($\tau=1$ for inverse exponential). The parameter $\tau$ is the shape parameter, which comes from the exponent $1 / \tau$. The scale parameter $\theta$ is added after raising the base distribution to a power.

Let $X$ be the random variable for the base exponential distribution. The following shows the information on the base exponential distribution.

Base Exponential
Density Function $f_X(x)=e^{-x} \ \ \ \ \ \ \ \ \ \ x>0$
CDF $F_X(x)=1-e^{-x} \ \ \ \ x>0$
Survival Function $S_X(x)=e^{-x} \ \ \ \ \ \ \ \ \ \ x>0$

Note that the above density function and CDF do not have the scale parameter. Once the base distribution is raised to a power, the scale parameter will be added to the newly created distribution.

The following gives the CDF and the density function of the transformed exponential distribution. The density function is obtained by taking the derivative of the CDF.

\displaystyle \begin{aligned} F_Y(y)&=P(Y \le y) \\&=P(X^{1 / \tau} \le y) \\&=P(X \le y^\tau)\\&=F_X(y^\tau) \\&=1-e^{- y^\tau} \ \ \ \ \ \ \ \ \ \ \ \ \ y>0 \end{aligned}

$\displaystyle f_Y(y)=\tau \ y^{\tau-1} \ e^{- y^\tau} \ \ \ \ \ \ \ \ \ y>0$

The following gives the CDF and the density function of the inverse exponential distribution.

\displaystyle \begin{aligned} F_Y(y)&=P(Y \le y) \\&=P(X^{-1} \le y) \\&=P(X \ge 1/y)\\&=S_X(1/y) \\&=e^{- 1/y} \ \ \ \ \ \ \ \ \ \ \ \ \ y>0 \end{aligned}

$\displaystyle f_Y(y)=\frac{1}{y^2} \ e^{- 1/y} \ \ \ \ \ \ \ \ \ y>0$

The following gives the CDF and the density function of the inverse transformed exponential distribution.

\displaystyle \begin{aligned} F_Y(y)&=P(Y \le y) \\&=P(X^{- 1 / \tau} \le y) \\&=P(X \ge y^{- \tau})\\&=S_X(y^{- \tau}) \\&=e^{- y^{- \tau}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y>0 \end{aligned}

$\displaystyle f_Y(y)= \frac{\tau}{y^{\tau+1}} \ e^{- 1/y^\tau} \ \ \ \ \ \ \ \ \ y>0$

The above derivation does not involve the scale parameter. Now it is added to the results.

Transformed Distribution
Transformed Exponential
CDF $F_Y(y)=1-e^{- (y/\theta)^\tau}$ $y>0$
Survival Function $S_Y(y)=e^{- (y/\theta)^\tau}$ $y>0$
Density Function $f_Y(y)=(\tau / \theta) \ (y/\theta)^{\tau-1} \ e^{- (y/\theta)^\tau}$ $y>0$
Inverse Exponential
CDF $F_Y(y)=e^{- \theta/y}$ $y>0$
Survival Function $S_Y(y)=1-e^{- \theta/y}$ $y>0$
Density Function $f_Y(y)=\frac{\theta}{y^2} \ e^{- \theta/y}$ $y>0$
Inverse Transformed Exponential
CDF $F_Y(y)=e^{- (\theta/y)^{\tau}}$ $y>0$
Survival Function $S_Y(y)=1-e^{- (\theta/y)^{\tau}}$ $y>0$
Density Function $f_Y(y)=\tau ( \theta / y )^\tau \ (1/y) \ e^{- (\theta/y)^{\tau}}$ $y>0$

The transformed exponential distribution and the inverse transformed distribution have two parameters $\tau$ and $\theta$. The inverse exponential distribution has only one parameter $\theta$. The parameter $\theta$ is the scale parameter. The parameter $\tau$, when there is one, is the shape parameter and it comes from the exponent when the exponential is raised to a power.

The above transformation starts with the exponential distribution with mean 1 (without the scale parameter) and the scale parameter $\theta$ is added back in at the end. We can also accomplish the same result by starting with an exponential variable $X$ with mean (scale parameter) $\theta^\tau$. Then raising $X$ to $1/\tau$, -1, and $-1/\tau$ would generate the three distributions described in the above table. In this process, the scale parameter $\theta$ is baked into the base distribution. This makes it easier to obtain the moments of the “transformed” exponential distributions since the moments would be derived from exponential moments.
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Connection with Weibull Distribution

Compare the density function for the transformed exponential distribution with the density of the Weibull distribution discussed here. Note that the two are identical. Thus raising an exponential distribution to $1 / \tau$ where $\tau >0$ produces a Weibull distribution.

On the other hand, raising a Weillbull distribution to -1 produces an inverse Weillbull distribution (by definition). Let $F_X(x)=1-e^{- x^\tau}$ be the CDF of the base Weibull distribution where $\tau >0$. Let’s find the CDF of $Y=X^{-1}$. Then add the scale parameter.

$\displaystyle F_Y(y)=P(Y \le y)=P(X \ge 1 / y)=e^{- (1/y)^\tau}$

$\displaystyle F_Y(y)=e^{- (\theta/y)^\tau}$ (scale parameter added)

Note the the CDF of inverse Weibull distribution is identical to the one for inverse transformed exponential distribution. Thus transformed exponential distribution is identical to a Weibull distribution and inverse transformed exponential distribution is identical to an inverse Weibull distribution.

Since Weibull distribution is the same as transformed exponential distribution, the previous post on Weibull distribution can inform us on transformed exponential distribution. For example, assuming that the Weibull distribution (or transformed exponential) is a model for the time until death of a life, varying the shape parameter $\tau$ yields different mortality patterns. The following are two graphics from the proevious post.

Figure 1

Figure 2

Figure 1 shows the Weibull density functions for different values of the shape parameter (the scale parameter $\theta$ is fixed at 1). The curve for $\tau=1$ is the exponential density curve. It is clear that the green density curve ($\tau=2$) approaches the x-axis at a faster rate then the other two curves and thus has a lighter tail than the other two density curves. In general, the Weibull (transformed exponential) distribution with shape parameter $\tau >1$ has a lighter tail than the Weibull with shape parameter $0<\tau <1$.

Figure 2 shows the failure rates for the Weibull (transformed exponential) distributions with the same three values of $\tau$. Note that the failure rate for $\tau=0.5$ (blue) decreases over time and the failure rate for $\tau=2$ increases over time. The failure rate for $\tau=1$ is constant since it is the exponential distribution.

What is being displayed in Figure 2 describes a general pattern. When the shape parameter is $0<\tau<0.5$, the failure rate decreases as time increases and the Weibull (transformed exponential) distribution is a model for infant mortality, or early-life failures. Hence these Weibull distributions have a thicker tail as shown in Figure 1.

When the shape parameter is $\tau >1$, the failure rate increases as time increases and the Weibull (transformed exponential) distribution is a model for wear-out failures. As times go by, the lives are fatigued and “die off.” Hence these Weibull distributions have a lighter tail as shown in Figure 1.

When $\tau=1$, the resulting Weibull (transformed exponential) distribution is exponential. The failure rate is constant and it is a model for random failures (failures that are independent of age).

Thus the transformed exponential family has a great deal of flexibility for modeling the failures of objects (machines, devices).

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Moments and Other Distributional Quantities

The moments for the three “transformed” exponential distributions are based on the gamma function. The two inverse distributions have limited moments. Since the transformed exponential distribution is identical to Weibull, its moments are identical to that of the Weibull distribution. The moments of the “transformed” exponential distributions are $E(Y)=E(X^{1 / \tau})$ where $X$ has an exponential distribution with mean (scale parameter) $\theta^\tau$. See here for the information on exponential moments. The following shows the moments of the “transformed” exponential distributions.

Name of Distribution Moment
Transformed Exponential $E(Y^k)=\theta^k \Gamma(1+k/\tau)$ $k >- \tau$
Inverse Exponential $E(Y^k)=\theta^k \Gamma(1-k)$ $k <1$
Inverse Transformed Exponential $E(Y^k)=\theta^k \Gamma(1-k/\tau)$ $k <\tau$

The function $\Gamma(\cdot)$ is the Gamma function. The transformed exponential moment $E(Y^k)$ exists for all $k >- \tau$. The moments are limited for the other two distributions. The first moment $E(Y)$ does not exist for the inverse exponential distribution. The inverse transformed exponential moment $E(Y^k)$ exist only for $k<\tau$. Thus the inverse transformed exponential mean and variance exist only if the shape parameter $\tau$ is larger than 2.

The distributional quantities that are based on moments can be calculated (e.g. variance, skewness and kurtosis) when the moments are available. For all three "transformed" exponential distributions, percentiles are easily computed since the CDFs contain only one instance of the unknown $y$. The following gives the mode of the three distributions.

Name of Distribution Mode
Transformed Exponential $\displaystyle \theta \biggl(\frac{\tau-1}{\tau} \biggr)^{1/\tau}$ for $\tau >1$, else 0
Inverse Exponential $\theta / 2$
Inverse Transformed Exponential $\displaystyle \theta \biggl(\frac{\tau}{\tau+1} \biggr)^{1/\tau}$

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$\copyright$ 2017 – Dan Ma