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Consider $n$ Bernoulli trials, where for $i=1,2,\dots,n$ the $i-$th trial has probability $p_i$ of success, and let $X$ be the random variable denoting the total number of successes. Let $p\ge p_i$ for all $i=1,2,\dots,n$. Prove that for $1≤k≤n$, $$\Pr(X < k) \ge \sum_{i=0}^{k-1}\dbinom{n}{i}p^i(1-p)^{(n-i)}$$

I looked for solutions and found an approach to find a common probability space and relate the two random variables. But the approach isn't clear to me. Here is the link: An inequality about binomial distribution

PS: this is no homework or assignment problem for me.

Robin
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GameOfChess
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2 Answers2

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In this situation $r_i\in[0,1]$ exists with $pr_i=p_i$.

Let $Y_1,\dots, Y_n$ be iid with Bernouilli distritution with parameter $p$.

Let $U_1,\dots, U_n$ be independent and let $U_i$ have Bernouilli distribution with parameter $r_i$.

Also let it be that $Y_i$ and $U_i$ are independent for each $i$.

Now define $X_i=Y_iU_i$ for $i=1,\dots,n$ and observe that $X_i\leq Y_i$.

Then $X_1,\dots, X_n$ are independent and $X_i$ has Bernouilli distribution with parameter $pr_i=p_i$.

Define $X=X_1+\cdots+X_n$ and $Y=Y_1+\cdots+Y_n$.

Then $Y$ has binomial distribution with parameters $n,p$,

From $X\leq Y$ it follows directly that: $$\Pr(X<k)\geq\Pr(Y<k)=\sum_{i=0}^{k-1}\binom{n}{i}p^i(1-p)^{n-i}$$

drhab
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An intuitive/non-rigorous answer is: we know that $p$ overestimates the probability of a successful trial, so if we have $Y \thicksim Bin(n, p)$, then for any $k$, $\mathbb{P}(Y < k) = \sum_{i=1}^{k-1} {n \choose i} p^i (1-p)^i$, which, since we are overestimating the likelihood of us succeeding $\mathbb{P}(X < k) \geq \mathbb{P}(Y < k)$.

Rigorous(ish)ly. Let $p \geq p_i$ $\forall i = 1, \ldots, n$, where $p_i$ denotes the probability of a success in the $i^{th}$ bernoulli trial. Then, let $X_1, X_2, \ldots, X_n$ be the $n$ Bernoulli trials that are not necessarily independent, and let $X = X_1 + X_2 + \ldots + X_n$. Then, let $Y \thicksim Bin(n,p)$. Finally: $$\mathbb{P}(X \geq k) = \mathbb{P}(\sum_{k=1}^n X_i \geq k) \leq \mathbb{P}(Y \geq k)$$ Which follows as every trial of $Y$ is more likely to succeed then every trial of $X_i$. Next, $$\mathbb{P}(X < k) = 1 - \mathbb{P}(X \geq k) \geq 1 - \mathbb{P}(Y \geq k) = \mathbb{P}(Y < k)\\\mathbb{P}(X < k) \geq \mathbb{P}(Y < k)$$ Which is equivalent to the above inequality.

Robert Murray
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