Hoeffding inequality for the difference of two sample means?

[JanO]

In W. Hoeffding's 1963 paper^* he gives the well known inequality:

P(\bar{x}-\mathrm{E}[x_i] \geq t) \leq \exp(-2t^2n) \ \ \ \ \ \ (1),

where \bar{x} = \frac{1}{n}\sum_{i=1}^nx_i, x_i\in[0,1]. x_i's are independent.

Following this theorem he gives a corollary for the difference of two sample means as:

P(\bar{x}-\bar{y}-(\mathrm{E}[x_i] - \mathrm{E}[y_k]) \geq t) \leq \exp(\frac{-2t^2}{m^{-1}+n^{-1}}) \ \ \ \ \ \ (2),

where \bar{x} = \frac{1}{n}\sum_{i=1}^nx_i, \bar{y} = \frac{1}{m}\sum_{k=1}^my_k, x_i,y_k\in[0,1]. x_i's and y_k's are independent.


My question is: How does (2) follow from (1)?

-Jan

^*http://www.csee.umbc.edu/~lomonaco/f08/643/hwk643/Hoeffding.pdf (equations (2.6) and (2.7))

 

[chiro]

Hey JanO and welcome to the forums.

One idea I have is to let Z = X + Y and use Z instead of X in the definition.

 

[JanO]

Thanks Chiro for your responce.

However, I still do not understand how the term (m^{-1} + n^{-1}) comes into the bound. Isn't z=\bar{x}-\bar{y} is still bounded between [0,1]?

-Jan

 

[chiro]

Thanks Chiro for your responce.

However, I still do not understand how the term (m^{-1} + n^{-1}) comes into the bound. Isn't z=\bar{x}-\bar{y} is still bounded between [0,1]?

-Jan

Think about what happens to the variances.

 

[JanO]

It seems like bounded here means all most surely bounded. At least that's how Hoeffding inequality seems to be given elsewhere. I guess it then means that z=\bar{x}-\bar{y} is bounded a.s. between [\mu_x-\mu_y-\frac{1}{2}\sqrt{m^{-1}+n^{-1}}, \ \mu_x-\mu_y+\frac{1}{2}\sqrt{m^{-1}+n^{-1}}].?

Thanks again for your help!