Help with a mathematical statistics question!

[euroazn]

Can somebody help with 7B?
http://www-stat.wharton.upenn.edu/~zongming/teaching/stat431/hw/hw1.pdf

Thanks

 

[weiwern]

Random guess here, since we're fixing the theoretical to the case with parameter \(\lambda=1\). Knowing that the mean of an exponential is \(\frac{1}{\lambda}\), we find the mean of our data, \(\bar{X}=\frac{1}{\hat{\lambda}}\rightarrow \hat{\lambda}=\frac{1}{\bar{X}}\). Using the hint provided, we just rescale our data to exponential with parameter 1 by multiplying \(\hat{\lambda}\) to our data (i.e. \(\hat{\lambda}x\) ~ Exp(1))

 

[euroazn]

Not quite sure what you mean; lambda is 1. So what you're saying is keep the original data, which is wrong :3

 

[weiwern]

Hmm, our theoretical distribution has lambda 1. And we're making the assumption that our data x1,...xn is also exponentially distributed but with unknown lambda. One way we can estimate the parameter of the distribution of our data is to get the sample mean which is equivalent to 1/lambda and we can retrieve our estimate of lambda from there. Now we have to rescale it by applying lambda to x which makes it exp(1) to match our theoretical distribution thus allowing us to compare the plotted points on y=x.

Okay, I guess a numerical example helps to shed some light into this. Imagine we have \((x_1,...,x_4)=(1,2,3,4)\) i.i.d. \( Exp(\lambda)\). So \(\bar{X}=2.5 \) which means \( {\lambda}=0.4\) so multiplying \(\lambda=0.4\) to our data we get \((\lambda x_1,..., \lambda x_4)=(0.4,0.8,1.2,1.6)\) which now has mean 1 or likewise \(\lambda = 1\). Hence, this allows us to compare it with our theoretical distribution of Exp(1)

 

[euroazn]

Essentially what you're saying then, is divide by the mean.
Unfortunately, this cannot be right - suppose the mean was 0!
I was thinking to transform into a normal distribution and then find z-scores, but it seems rather complicated :/

 

[pruse]

weiwern is right...

according to the hint, if we multiply the data by lambda, we get the rescaling we want. but lambda is unknown, so we instead multiply by an estimate of lambda, which for the exponential distribution is the reciprocal of the sample mean.

the mean cannot be zero because our data are presumably always positive -- or else it would be unreasonable to test for an exponential distribution in the first place.

 

[weiwern]

Thanks for the clarification, peterruse. Sometimes I get boggled by the maths that my reasoning suffers from it. The only way you'll have mean of 0 is all your x's were 0 and that would be absurd to model it as exponentially distributed. Support of exponential distribution is [0, infinity)