Show $P(\frac{1}{n}\sum_{i=1}^{n}Y_i\geq c)\leq e^{-nd}$ for constants $c$ and $d$ Let $Y_1, Y_2\ldots$ be a sequence of i.id. random variables uniformly distributed on $[0,1]$. Let $c>\frac{1}{2}$. Show that there exists $d>0$ (depends on $c$) such that 
$$P\left(\frac{1}{n}\sum_{i=1}^{n}Y_i\geq c\right)\leq e^{-nd}$$
What I could do best so far:
$E(Y_i)=\frac{1}{2}$ and $V(Y_i)=\frac{1}{12}$ since $Y_i$ are uniformly distributed on $[0,1]$.
By the Weak Law of Large Numbers:
$$P\left(\Big|\frac{1}{n}\sum_{i=1}^{n}(X_i-E(X_i))\Big|\geq\epsilon\right)\leq\frac{V(X_i)}{n\epsilon^2}$$
then
$$P\left(\Big|\frac{1}{n}\sum_{i=1}^{n}X_i-\sum_{i=1}^{n}E(X_i)\Big|\geq\epsilon\right)\leq\frac{V(X_i)}{n\epsilon^2}$$
$$P\left(\Big|\frac{1}{n}\sum_{i=1}^{n}X_i-\frac{n}{2}\Big|\geq\epsilon\right)\leq\frac{1}{12n\epsilon^2}$$
I am stuck here.
I have little clue how to get the exponential function.
Could someone please give some light on how to approach and proceed with the problem?
Thanks.
EDITED
Thanks @Fnacool for the clue. But I was still stuck in some steps:
$$P\left(\frac{1}{n}\sum_{i=1}^{n}Y_i\geq c\right)=P\left(e^{\theta\sum_{i=1}^{n}Y_i}\geq e^{\theta cn}\right)\leq\frac{E\left(e^{\theta\sum_{i=1}^{n}Y_i}\right)}{e^{\theta cn}}=\frac{E[e^{\theta Y}]^n}{e^{\theta cn}}$$
Now $E[e^{\theta Y}]^n=(\int_0^1 e^{\theta y}\times1 dy)^n$ since pdf of $Y$ is $f(y)=1$
Then 
$$\left(\int_0^1 e^{\theta y}\times1 dy\right)^n=\left(\left[\frac{1}{\theta}e^{\theta y}\right]_0^1\right)^n=\left(\frac{e^{\theta}}{\theta}-\frac{1}{\theta}\right)^n$$
Then I was stuck here. How can we perform the multinomial expansion?
 A: Clue: 


*

*$\{\sum_{i=1}^n Y_i > cn\} = \{e^{\theta \sum_{i=1}^n Y_i }>e^{\theta cn}\}$ for any $\theta>0$.  

*For positive random variable, $Z$ and $z>0$, $P(Z>z)\le E[Z]/z$. 

*$E[e^{\theta \sum_{i=1}^n Y_i}] = E[e^{\theta Y_1}]^n$

*Next, 


$$ E [ e^{\theta Y}]= \int_0^1 e^{\theta u} du = \frac{e^{\theta}-1}{\theta}=e^{\theta} \left( \frac{1-e^{-\theta}}{\theta}\right).$$
Therefore, our upper bound is 
$$ e^{n \theta}  \left( \frac{1-e^{-\theta}}{\theta}\right)^n e^{-\theta c n}.$$
Take logarithm to obtain 
$$ n \left ( \theta (1-c)+\ln (1-e^{-\theta})-\ln \theta\right)=(*)$$ 
We need this to be negative. We will optimize over $\theta$. In principle, small enough $\theta$ will do the trick. Here is a bare hands approach.
Take $e^{-\theta} > 1-\theta+ \theta^2/2 -\theta^3/6$ (fourth derivative is positive). Therefore, 
$$\ln (1-e^{-\theta}) < \ln \theta + \ln (1-\theta/2+\theta^2/6) .$$
and then 
$$(*) < n (\theta(1-c)+ \ln (1-\theta/2+\theta^2/6)).$$
Using $\ln (1-x)<-x$ for $x<1$ and choosing $\theta$ so that $\theta/2-\theta^2/6<1$, we finally have 
$$ (*) < n (\theta (1-c) -\theta/2+\theta^2/6)=n\theta( (\frac 12 -c) + \frac{\theta}{6}).$$ 
For small enough $\theta>0$ this is negative.  
