# Upper bound on tail probability $P(X \geq 100)$ with given generating function at a point, $\mathbb E (10^X)=10$

Find the best possible upper bound on $$P(X \geq 100)$$ if $$g_X(10)=10$$ ($$g_X$$ is a generating function), where $$X$$ is a discrete random variable.

I started by writing the equation to find possible generating function $$g_X(10)=(1-p)\cdot10^0+p\cdot10^{100} = 10$$.
Its solution is $$p = \frac{9}{10^{100}-1}$$, hence $$P(X\geq 100) = \frac{9}{10^{100}-1}$$.

Is it possible to give a better estimate?
It seems to me that it is not, because as we increase the exponent at $$p\cdot10^{100}$$, (e.g. $$p\cdot10^{101}$$) , then $$p$$ decreases, and as we increase the exponent at ($$1-p)\cdot10^0$$ (e.g. ($$1-p)\cdot10^1$$), then $$p$$ is zero or negative.
Although I'm not sure, which is why I'm asking.

• Are you assuming $X$ is a Bernoulli RV? Feb 18 at 15:27
• @GiorgosGiapitzakis It's a discrete random variable, not necessarily one with Bernoulli distribution. Feb 18 at 16:03

From the OP, I guess you assumed that $$X$$ is non-negative, which implies $$10^X-1 \ge 0.$$

Now, from the Markov's inequality we have

$$\mathbb P (X \geq 100)=\mathbb P (10^X-1 \geq 10^{100}-1) \le \frac{\mathbb E (10^X-1)}{10^{100}-1}=\frac{g_X(10)-1}{10^{100}-1}=\color{blue}{\frac{9}{10^{100}-1}}$$

where recall $$g_X(t):=\mathbb E (t^X)$$ for $$t>0$$. This gives the bound you are seeking, which is tight for a random variable $$X_0$$ with

$$\mathbb P (10^{X_0}-1=10^{100}-1)=\frac{9}{10^{100}-1},\mathbb P (10^{X_0}-1=0)=1-\frac{9}{10^{100}-1}.$$

If $$X$$ can take negative values, then

$$10^X > 0.$$

In this case, again from the Markov's inequality we have

$$\mathbb P (X \geq 100)=\mathbb P (10^X \geq 10^{100}) \le \frac{\mathbb E (10^X)}{10^{100}}=\frac{g_X(10)}{10^{100}}=\color{blue}{\frac{10}{10^{100}}}.$$

This bound is not tight, but can be approached arbitrarily by the sequence $$X_n$$ defined as ($$n \in \mathbb N$$)

$$\mathbb P (10^X-10^{-n}=10^{100}-10^{-n})=\frac{10-10^{-n}}{10^{100}-10^{-n}},\mathbb P (10^X-10^{-n}=0)=1-\frac{10-10^{-n}}{10^{100}-10^{-n}},$$

for which we have

$$g_{X_n}(10)=10, \mathbb P (X_n \geq 100) = \frac{10-10^{-n}}{10^{100}-10^{-n}} <\frac{10}{10^{100}}.$$

You can see that $$\frac{10-10^{-n}}{10^{100}-10^{-n}}$$ is increasing in $$n$$ and tends to the bound $$\frac{10}{10^{100}}$$ when $$n \to \infty$$.

• I do not assume anything about $X$, except that it is a discrete random variable. In other words, X can be negative. Does this change the estimate? In my case $X$ takes $0$ or $100$, just because that seemed the best estimate to me. I once saw a similar question in which there was a hint to use $g_X(t)=E[t^X]$, but I don't know the solution. Also, thanks for your answer. Feb 26 at 18:43
• @Michał In the updated answer, I considered the case where $X$ can be negative. You may see that the new bound is greater than the bound obtained for the case where $X$ is non-negative, and it is not tight anymore.
– Amir
Feb 26 at 19:37