Exponential Laws as Special Gamma Cases

One Distribution, Two Familiar Forms

The exponential distribution and the gamma distribution are often introduced as separate models. Their density formulas show a direct connection: an exponential law is precisely a gamma law whose shape parameter is one. Recognizing this special case lets us move between the two parameterizations without changing the distribution, and provides a useful check on formulas for gamma densities and moments.

The key is to keep the parameters distinct. We write the exponential rate as $\lambda$ and its scale as $\theta$, with $\theta=1/\lambda$. For a gamma law, we use shape $\alpha$ and either rate $\lambda$ or scale $\theta$. Setting $\alpha=1$ gives the exponential law, with the same rate or scale, respectively.

Definitions and Parameter Matching

Definition: For $\lambda>0$, an exponential random variable with rate $\lambda$ has density $$ f(x)=\lambda e^{-\lambda x},\qquad x>0, $$ and density zero otherwise. Its scale is $\theta=1/\lambda$, so the same density can be written $f(x)=\theta^{-1}e^{-x/\theta}$ for $x>0$. A gamma random variable with shape $\alpha>0$ and rate $\lambda>0$ has density $$ f(x)=\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x},\qquad x>0, $$ and density zero otherwise, where $\Gamma(\alpha)=\int_0^\infty u^{\alpha-1}e^{-u}\,du$. In the scale parameterization, with $\theta=1/\lambda$, its density is $f(x)=x^{\alpha-1}e^{-x/\theta}/\{\Gamma(\alpha)\theta^\alpha\}$ for $x>0$.

The gamma function satisfies $\Gamma(1)=1$, since $\int_0^\infty e^{-u}\,du=1$. Thus, when $\alpha=1$, the factor $x^{\alpha-1}$ is $1$, and the gamma density simplifies to $\lambda e^{-\lambda x}$. In the scale parameterization, $\theta^{-1}e^{-x/\theta}$ is the same density. The support is $x>0$ in both cases.

Theorem: A gamma distribution with shape $\alpha=1$ and rate $\lambda$ is the exponential distribution with rate $\lambda$. Equivalently, a gamma distribution with shape $\alpha=1$ and scale $\theta$ is the exponential distribution with scale $\theta$, where $\theta=1/\lambda$.

Proof. Substitute $\alpha=1$ into the gamma rate density. Since $x^0=1$ for $x>0$ and $\Gamma(1)=1$,

\frac{\lambda^1}{\Gamma(1)}x^{1-1}e^{-\lambda x} =\lambda e^{-\lambda x},\qquad x>0.

This is the exponential density with rate $\lambda$. Replacing $\lambda$ by $1/\theta$ gives $\theta^{-1}e^{-x/\theta}$, the exponential density with scale $\theta$. Hence the two parameterizations describe the same distribution. $\square$

Worked Example: Exponential Rate as Gamma Rate

Suppose $X$ is exponential with rate $\lambda=2.5$. Express its distribution as a gamma law and verify the density and moments.

The rate-to-scale conversion gives $\theta=1/2.5=0.4$. The same variable is gamma with shape $\alpha=1$ and rate $2.5$. Its gamma density is

f(x)=\frac{2.5^1}{\Gamma(1)}x^0e^{-2.5x} =2.5e^{-2.5x},\qquad x>0.

This is the exponential density with rate $2.5$. The exponential mean and variance are $1/\lambda=0.4$ and $1/\lambda^2=0.16$. The gamma formulas give $E[X]=\alpha/\lambda=1/2.5=0.4$ and $\operatorname{Var}(X)=\alpha/\lambda^2=1/2.5^2=0.16$, confirming the match.

Why the Moments Agree

The rate-scale correspondence is not just a density identity: the gamma moment formula reduces to the exponential moment formula when the shape is one. We derive the relevant formula to make the substitution explicit. For a gamma variable with shape $\alpha>0$ and rate $\lambda>0$, the raw moment of order $r$ exists when $r>-\alpha$.

Result: If $X$ has a gamma distribution with shape $\alpha>0$ and rate $\lambda>0$, then, for $r>-\alpha$, $$ E[X^r]=\frac{\Gamma(\alpha+r)}{\Gamma(\alpha)\lambda^r}. $$ In particular, for $\alpha=1$ and $r>-1$, $E[X^r]=\Gamma(r+1)/\lambda^r$. Therefore $E[X]=1/\lambda$ and $\operatorname{Var}(X)=1/\lambda^2$.

Derivation. By the definition of expectation and the gamma density,

E[X^r] =\int_0^\infty x^r\frac{\lambda^\alpha}{\Gamma(\alpha)} x^{\alpha-1}e^{-\lambda x}\,dx =\frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha+r-1}e^{-\lambda x}\,dx.

The condition $r>-\alpha$ ensures that the corresponding gamma integral is finite near zero; the exponential factor ensures convergence at infinity. In the integral, set $u=\lambda x$, so $x=u/\lambda$ and $dx=du/\lambda$. It follows that

E[X^r] =\frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^\infty \left(\frac{u}{\lambda}\right)^{\alpha+r-1} e^{-u}\frac{du}{\lambda} =\frac{\Gamma(\alpha+r)}{\Gamma(\alpha)\lambda^r}.

For the mean and variance at shape one, the formula gives $E[X]=\Gamma(2)/\{\Gamma(1)\lambda\}=1/\lambda$, and $E[X^2]=\Gamma(3)/\{\Gamma(1)\lambda^2\}=2/\lambda^2$. Here $\Gamma(2)=1$ and $\Gamma(3)=2$, obtained from the gamma recurrence $\Gamma(t+1)=t\Gamma(t)$ for $t>0$. Consequently,

\operatorname{Var}(X)=E[X^2]-(E[X])^2 =\frac{2}{\lambda^2}-\frac{1}{\lambda^2} =\frac{1}{\lambda^2}.

In scale form, substituting $\lambda=1/\theta$ yields $E[X]=\theta$ and $\operatorname{Var}(X)=\theta^2$, as required for the exponential scale parameterization. $\square$

Worked Example: Gamma Scale as Exponential Scale

Let $Y$ have a gamma distribution with shape $\alpha=1$ and scale $\theta=3$. Find its exponential parameterization and check its first two moments.

The corresponding rate is $\lambda=1/\theta=1/3$. Substitution into the gamma scale density gives

f(y)=\frac{y^{1-1}e^{-y/3}}{\Gamma(1)3^1} =\frac{1}{3}e^{-y/3},\qquad y>0.

Thus $Y$ is exponential with scale $3$, or equivalently with rate $1/3$. Its mean is $1/\lambda=3$, and its variance is $1/\lambda^2=9$. Directly from the gamma scale formulas, $E[Y]=\alpha\theta=1(3)=3$ and $\operatorname{Var}(Y)=\alpha\theta^2=1(3^2)=9$. Both the density and moments agree.

The Converse: When Can a Gamma Law Be Exponential?

The shape-one condition is also necessary: no gamma distribution with a different shape can have the same distribution as an exponential variable, even if the two rates are allowed to differ. It matters to allow different rates in this comparison; matching only the gamma and exponential rates would not establish the full claim.

Theorem: Let $X$ have a gamma distribution with shape $\alpha>0$ and rate $\lambda>0$, and let $Z$ have an exponential distribution with rate $\mu>0$. Their distributions are equal if and only if $\alpha=1$ and $\lambda=\mu$.

Proof. Both densities are continuous and strictly positive on $x>0$. If the distributions are equal, their densities agree almost everywhere; continuity then implies that they agree at every $x>0$. The ratio of the gamma density to the exponential density is therefore identically one:

\frac{\lambda^\alpha x^{\alpha-1}e^{-\lambda x}/\Gamma(\alpha)} {\mu e^{-\mu x}} = \frac{\lambda^\alpha}{\Gamma(\alpha)\mu} x^{\alpha-1}e^{-(\lambda-\mu)x} =1,\qquad x>0.

Taking the logarithm and differentiating with respect to $x$ is valid because the ratio is positive and differentiable for $x>0$. Since it is constant, its logarithmic derivative must vanish for every $x>0$:

\frac{\alpha-1}{x}-(\lambda-\mu)=0,\qquad x>0.

Multiplying by $x$ gives $\alpha-1-(\lambda-\mu)x=0$ for every $x>0$. This expression can be identically zero only if its constant coefficient is zero and its coefficient of $x$ is zero. Hence $\alpha=1$ and $\lambda=\mu$. Conversely, these parameter values make the densities identical by the shape-one calculation above. $\square$

Worked Example: A Gamma Density That Cannot Be Exponential

Consider a gamma variable with shape $\alpha=2$ and rate $\lambda=3$. Check whether it could equal an exponential distribution with some rate $\mu>0$, not necessarily $3$.

Since $\Gamma(2)=1$, the gamma density is $9xe^{-3x}$ for $x>0$. Dividing by the density $\mu e^{-\mu x}$ of a proposed exponential law gives the ratio

R(x)=\frac{9}{\mu}x e^{-(3-\mu)x}.

For the densities to be equal, this ratio would have to equal one for every $x>0$, so its logarithmic derivative would have to vanish for every such $x$. But

\frac{d}{dx}\log R(x)=\frac{1}{x}-(3-\mu).

At $x=1$, this derivative is zero only if $\mu=2$. At $x=2$, it is zero only if $\mu=2.5$. No single rate satisfies both conditions, so the gamma distribution with shape two cannot be exponential at any rate. This specific check illustrates why the converse comparison must allow the exponential rate to vary.

Common Mistake and Practical Check

A frequent error is to set $\alpha=1$ but leave the gamma rate and exponential scale numerically equal. They are reciprocal when one is a rate and the other is a scale. For example, a gamma shape-one law with rate $4$ is exponential with rate $4$, or with scale $1/4$—not with scale $4$. A reliable check is to write both densities on the same parameter scale before comparing them.

This special case is useful when a calculation is stated in one family but is easier to interpret in the other. The gamma notation makes the exponential law part of a broader shape-and-rate family; the exponential notation highlights the single rate or scale that determines its behavior. In either form, matching the density guarantees matching probabilities and moments.

Key takeaway: Exponential $(\lambda)$ is Gamma $(\alpha=1,\text{ rate } \lambda)$, equivalently Gamma $(\alpha=1,\text{ scale }1/\lambda)$. Conversely, a gamma distribution can equal an exponential distribution only when its shape is one and its rate matches the exponential rate.

Check Your Understanding

Use the density and parameter conversions to answer each question.

A gamma variable has shape one and rate $5$. What are its exponential rate and scale?
A gamma variable has shape one and scale $0.25$. What is its exponential rate?
For an exponential variable with rate $2$, calculate the mean and variance using the gamma special case.
Why does comparing only gamma and exponential densities with the same rate fail to prove the converse theorem?
Can a gamma distribution with shape $3$ equal an exponential distribution with a different rate? State the condition in the converse theorem that decides this.