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Suppose $Y_1 \dots Y_r$ are exponentially distributed i.i.d. RVs with parameter $\lambda$. I am supposed to show that for $n>r$ and

$$T_r = Y_1 + \dots + Y_r + (n-r)Y_r$$

the RV $2\lambda T_r$ is $\chi^2$ distributed with 2r degrees of freedom.

Here's what I've tried so far:

I tried using the transformation $(Y_1 \dots Y_r) \mapsto (Y_1 \dots Y_{r-1}, T_r)$ and evaluate the joint density $f_{Y_1, \dots ,Y_{r-1}, T_r}$ via the transformation formula. Integrating out $T_r$, this led me to

$$f_{T_r}(y)=\lambda \frac{(n-r+1)^{r-1}}{(n-r)^{r-1}}e^{-\lambda \frac{y}{n-r+1}}.$$

Adjusting for $2\lambda$ I get the density

$$f_{2\lambda T_r}(y)=1/2 \frac{(n-r+1)^{r-1}}{(n-r)^{r-1}}e^{-\lambda \frac{y}{2(n-r+1)}}.$$

However, I don't see how this is a $\chi^2$ distribution in any sense. What did I do wrong? What would be a better approach if this one doesn't work?

BabyWienerSpace
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  • The statement in your post is incorrect. What is true is that $$2\lambda \left{\sum_{i=1}^r Y_{(i)}+(n-r)Y_{(r)}\right} \sim \chi^2_{2r},,$$ when $Y_1,Y_2,\ldots,Y_n$ are i.i.d Exponential with rate $\lambda$. This can be shown using a standard transformation: https://math.stackexchange.com/q/4344872/321264. – StubbornAtom Jun 08 '23 at 07:21

1 Answers1

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We can use moment-generating functions to do the computation: let $$M_{T_r}(z) = \operatorname{E}[e^{T_r z}] \tag{1}$$ be the MGF of $T_r$. Then since the $Y_i$ are IID, $$\begin{align} M_{T_r}(z) &= \operatorname{E}[e^{Y_1 z} e^{Y_2 z} \cdots e^{Y_{r-1} z} e^{(n+1-r) Y_r z}] \\ &\overset{\text{iid}}{=} \operatorname{E}[e^{Y_1 z}] \operatorname{E}[e^{Y_2 z}] \cdots \operatorname{E}[Y_{Y_{r-1} z}] \operatorname{E}[e^{(n+1-r) Y_r z}] \\ &= M_{Y_1}(z) M_{Y_2}(z) \cdots M_{Y_{r-1}}(z) M_{Y_r}((n+1-r)z) \\ &\overset{\text{iid}}{=} (M_{Y_1}(z))^{r-1} M_{Y_1}((n+1-r)z). \tag{2} \end{align}$$

Now since the MGF of the exponential distribution with rate $\lambda$ is $$M_{Y_1}(z) = \operatorname{E}[e^{Y_1 z}] = \int_{y=0}^\infty e^{yz} \lambda e^{-\lambda y} \, dy = \frac{\lambda}{\lambda - z}, \quad \lambda > z > 0, \tag{3}$$ it follows that $$M_{T_r}(z) = \left(\frac{\lambda}{\lambda - z}\right)^{r-1} \left(\frac{\lambda}{\lambda - (n+1-r)z}\right) = \frac{\lambda^r}{(\lambda - z)^{r-1} (\lambda - (n+1-r)z)}. \tag{4}$$ Then $2\lambda T_r$ has MGF $$M_{2\lambda T_r}(z) = M_{T_r}(2\lambda z) = \frac{1}{(1-2z)^{r-1} (1-2(n+1-r)z)} \tag{5}$$ and it should now be clear that this cannot be $\chi^2$ distributed unless $n = r$, since its MGF has the form $$M_{\chi^2(2r)} (z) = \frac{1}{(1-2z)^r}. \tag{6}$$

We can also see why the claimed result cannot be true simply by asking where the value of $n$ went to. $2\lambda T_r$ will depend on $n$, yet it is claimed that it follows a distribution that does not depend on $n$.

heropup
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  • Thanks a lot! Initially I tried a similar approach by using characteristic functions which almost works out the same way. I came to the same conclusion, but my tutor told me that the statement was true anyways. Well, you never know. – BabyWienerSpace Jun 05 '23 at 19:07
  • @BabyWienerSpace Unless you have missstated the question in some way, the claim is incorrect, and it becomes your tutor's responsibility to prove otherwise. As I have already stated, $T_r$ as you have defined it, must depend on $n$. If $n$ is large relative to $r$, $2\lambda T_r$ will behave more like $2\lambda n Y_r$--namely, exponential with mean $2n$. If $n$ is small relative to $r$, then $2\lambda T_r$ will behave more like a chi-squared distribution. – heropup Jun 05 '23 at 20:48
  • Let me cite the task: "Suppose n components are tested and the experiment terminates as soon as r of them have failed. Assume component lifetimes are independent, each having an exponential distribution with parameter λ. Let Yi denote the time at which the ith failure occurs, so that Tr = Y1 + . . . + Yr + (n − r)Yr is the total accumulated lifetime at termination." Feel free to tell me if I neglected some crucial information. – BabyWienerSpace Jun 05 '23 at 21:53
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    @BabyWienerSpace From the description of the problem, there is no reason to believe that the $Y_i$ are themselves exponentially distributed. That is not what the problem states. The $Y_i$ are order statistics of the individual component lifetimes. For example, if $n = 4$ and $r = 2$, then let $X_1, X_2, X_3, X_4$ are the IID exponential lifetimes, and $Y_1 = \min_i X_i$ is the first order statistic. $Y_2$ is the second order statistic. – heropup Jun 06 '23 at 03:29