2
$\begingroup$

Let $X_1, \cdots, X_n$ be i.i.d. (positive) random variables with mean $1$ and density $f(x)$ (we can add conditions like sub-gaussian, sub-gamma later). Now I was interested in the following probability:

$$\mathbb P(|\frac{S_m} m -1|>\epsilon | S_n =n )$$

here $S_m = \sum_ {i =1}^ m X_m,$ $S_n = \sum_ {i =1}^ n X_n $.

Is there standard way to bound this probability? If we add conditions like sub-gaussian or sub-gamma on $X_1$, would it be possible to get sharper bound of this conditioned probability than the unconditioned probability?

In particular, I am not even sure if the following is true for general random variables or after adding conditions:

$$\mathbb P(|\frac{S_m} m -1|>\epsilon | \frac{S_n} n \in [1-\epsilon', 1+\epsilon'] ) \leq \mathbb P(|\frac{S_m} m -1|>\epsilon)$$ when $\epsilon'$ is small comparing with $\epsilon$. Intuitively this seems to make sense since after we know the sum is around its mean, the probability of its first few terms seems to have larger probability concentrate on its mean. However I have no idea how to prove this/ give counterexamples.

$\endgroup$
  • $\begingroup$ Just want to make sure if $m < n$ first, otherwise you can easily simplify it. $\endgroup$ – BGM Jun 21 '18 at 1:41
  • $\begingroup$ Yes here we have $m<n$ $\endgroup$ – Newo Jun 21 '18 at 2:18
  • $\begingroup$ Just a thought, if you know that $\sum_{j=1}^n X_j = n$, then $\sum_{j=1}^m X_j= n - \sum_{j=m+1}^n X_j$. So you need $\mathbb{P}(\vert \frac{n - \sum_{j=m+1}^n X_j}{m} -1 \vert > \epsilon)$ $\endgroup$ – blanchey Jun 21 '18 at 5:24
0
$\begingroup$

If you know that $\sum_{j=1}^n X_j = n$, then $\sum_{j=1}^m X_j= n - \sum_{j=m+1}^n X_j$. So you need $\mathbb{P}(\vert \frac{n - \sum_{j=m+1}^n X_j}{m} -1 \vert > \epsilon)$. Multiply through by m, getting you $\mathbb{P}(\vert S_{n-m} - (n-m) \vert > m\epsilon )$, and finish it off with Chebyshev's inequality

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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