Let's say that I have two exponential distributions with a mean of 10.
Now consider the mixture distribution where there is 50-50 chances of getting either one of the two exponential.
What would be the variance of this mixture distribution?
Asked
Active
Viewed 11 times
0
-
2The variance is the same as the variance of the exponential distribution(s) with mean $10$. The answer would be different if the two means were different. – Dilip Sarwate Mar 12 '20 at 16:37
-
@Dilip You implicitly assume the two exponentials have the same scale factor, whereas the question strongly implies that is not the case, for otherwise the situation would be trivial. – whuber Mar 12 '20 at 16:47
-
@whuber I don't understand what you mean by scale factor. Don't both exponential distributions have density $\mu^{-1}\exp(-\mu^{-1}x)\mathbb 1_{x\geq 0}$? And aren't the weights specified as 50-50? But even if they weren't, the mixture density would be $$(\alpha)\mu^{-1}\exp(-\mu^{-1}x)\mathbb 1_{x\geq 0}+(1-\alpha)\mu^{-1}\exp(-\mu^{-1}x)\mathbb 1_{x\geq 0} = \mu^{-1}\exp(-\mu^{-1}x)\mathbb 1_{x\geq 0}$$ leading to the same variance for the mixture density as the variances of the individual exponential random variables. Or, I am wildly mis-interpreting the question. – Dilip Sarwate Mar 12 '20 at 17:26
-
1@Dilip (As usual,) you're right: I was thinking of the Gamma family more generally. Your point is that the mean (in this case) determines the distribution. – whuber Mar 12 '20 at 17:32