Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X_1,\ldots,X_n$ be exponentially distributed with parameter $\lambda$ This implies that $Y=\sum_{i=1}^nX_i$ has a gamma distribution with parameters $(\lambda,n)$

Can anyone help me show that $$\left( \frac{a}{n\bar{x}}, \frac{b}{n\bar{x}} \right)$$ Is an exact $95$% central confidence interval for $\lambda$ if

$$\int_0^a \frac{y^{n-1}e^{-y}\;dy}{\Gamma(n)}= \int_b^\infty \frac{y^{n-1}e^{-y}\;dy}{\Gamma(n)}= 0.025$$

Here is what I have so far. Basically i've been trying to construct anything to help use those given integrals


$$=\lambda^{-n}\int_0^\frac{a}{n\bar{x}} \frac{y^{n-1}e^{-y/\lambda}\;dy}{\Gamma(n)}+\lambda^{-n}\int_0^\frac{b}{n\bar{x}} \frac{y^{n-1}e^{-y/\lambda}\;dy}{\Gamma(n)}-1$$

Which is where I get stuck as this doesn't really look salvagable. Any help here would be greatly appreciated!

share|improve this question

2 Answers 2

up vote 1 down vote accepted

You wrote $$P\left(\frac{a}{n\bar{x}}<Y<\frac{b}{n\bar{x}}\right).$$ But what you need is $$P\left(\frac{a}{n\bar{X}}<\lambda<\frac{b}{n\bar{X}}\right) = 0.95.\tag{1}$$ I've set $\bar X$ in capital since it's a random variable. One must remember what is random and what is not random in this kind of problem. To say that $a/(n\bar X)$ is "random" in effect means that if you take another sample, the value of that expression will change. $\lambda$ on the other hand is not random since it will remain the same if another sample is taken.

(1) is equivalent to $$ P\left(\frac a\lambda < n\bar X < \frac b\lambda\right) = 0.95, $$ or $$ P\left(\frac a\lambda < Y < \frac b\lambda\right) = 0.95 \tag{2} $$ Here is an ambiguity in the question: does "exponential with parameter $\lambda$" mean having density proportional to $y\mapsto e^{-\lambda y}$ or does it mean $y\mapsto e^{-y/\lambda}$? Since (2) is equivalent to $$ P\left(a < \lambda Y < b\right) = 0.95, $$ I take it to mean $\lambda Y$ has a gamma distribution with parameter $1$ in place of $\lambda$, so the density of the exponential is proportional to $y\mapsto e^{-\lambda y}$, i.e. $\lambda$ is an intensity parameter rather than a scale parameter.

So $$ P(X_i > c) = \int_c^\infty e^{-\lambda u} (\lambda \; du) = \int_{\lambda c}^\infty e^{-y}\;dy, $$ and so $$ P(Y>c) = (X_1 + \cdots + X_n > c) = \int_{\lambda c}^\infty \frac{y^{n-1} e^{-y}}{\Gamma(n)} \; dy. $$

Finally we have $$ P\left( \frac a\lambda < Y < \frac b\lambda \right) = \int_{\lambda(a/\lambda)}^{\lambda(a/\lambda)} \cdots\cdots = \int_a^b \cdots\cdots. $$

share|improve this answer
Thank you this is incredibly helpful! Thank you for being slow and clear in your answer, it made it very easy to understand –  Freeman Feb 8 '12 at 13:01

Assuming that $E[X_i] = \lambda^{-1}$, $Y = \sum_i X_i = n\bar{X}$ is a Gamma random variable with mean $n\lambda^{-1}$ and density function $$f_Y(y) = \lambda \frac{(\lambda y)^{n-1}}{\Gamma(n)}\exp(-\lambda y)\mathbf{1}_{(0,\infty)}.$$ Define $\alpha$ and $\beta$ as the solutions to $F_Y(\alpha)=0.025$ and $F_Y(\beta) = 0.975$ so that $$P\{\alpha \leq Y \leq \beta\} = F_Y(\beta) - F_Y(\alpha) = 0.95$$ and note that $$F_Y(\alpha) = \int_0^{\alpha} \lambda \frac{(\lambda y)^{n-1}}{\Gamma(n)}\exp(-\lambda y) \mathrm dy = \int_0^{\lambda\alpha} \frac{t^{n-1}}{\Gamma(n)}\exp(-t) \mathrm dt$$ so that $a = \lambda \alpha$ and similarly $b = \lambda \beta$. Verify for yourself that the following bounds hold: $$ \alpha < n\lambda^{-1} < \beta; ~~ a < n < b.$$

Now suppose that the value of $\lambda$ is unknown. We observe the values of the $X_i$, compute the value of $Y = \sum_i X_i = n\bar{X}$, and have $95\%$ confidence that $Y$ is in the interval $[\alpha, \beta]$ whose end-points are, unfortunately, unknown to us. But, if we assume that the known value of $Y$ is at the endpoint $\alpha = a/\lambda$, then we are in effect estimating that the unknown value of $\lambda$ is $a/Y = a/n\bar{X}$ while if we assume that $Y$ is at the other endpoint $\beta = b/\lambda$, then we are in effect estimating that the unknown value of $\lambda$ is $b/Y = b/n\bar{X}$. More generally, if we simply assume that $Y$ has taken on the expected value $n\lambda^{-1}$, we in effect estimate the value of $\lambda$ as $$\hat{\lambda} = \frac{n}{Y} = \frac{n}{X_1 + X_2 + \cdots + X_n}$$ which will be readily recognized as the maximum-likelihood estimate of the parameter $\lambda$ of an exponential random variable $X$ based on $n$ independent observations of $X$. But now we also have a $95\%$ confidence interval for our estimate.

If the sample mean is $\bar{X}$, then $\left(\frac{a}{n\bar{X}},\frac{b}{n\bar{X}}\right)$ is an exact $95\%$ confidence interval for the unknown parameter $\lambda$.

share|improve this answer
Thank you so much for this very detailed answer, it was very helpful. I feel I know understand this –  Freeman Feb 8 '12 at 13:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.