Assume we have two random variable $X,Y$ where X counts the number of ''heads'' in 100 tosses with probability $0<p_1<1$ and $Y$ counts the number of ''heads'' with probability $0<p_2<1$.
If $p_1<p_2$ prove that $P(X \geq 30) \leq P(Y \geq 30)$
We can exress $X=\sum_{i=1}^{100}X_i$ where $X_i=1$ if the coin indicates ''head'' and $0$ otherwise.
We do the exact same thing with $Y$.
I tried to use the binomial distribution to compute these two probabilities but i did not reach to a conclusion.
Can someone help me with this problem?
Thank you in advance.
For $X$ with binomial distribution $Binom(100,p)$ consider $$ f(p)=\mathsf P(X\geq 30)=\sum_{k=30}^{100}\binom{100}{k}p^k(1-p)^{100-k} $$ and prove that $f(p)$ increases when $p$ increases.
Find derivative $f'(p)$ $$ \frac{df(p)}{dp}=\sum_{k=30}^{100}\binom{100}{k}\cdot\frac{d}{dp}\left(p^k(1-p)^{100-k}\right) = \sum_{k=30}^{100}\binom{100}{k}\left(kp^{k-1}(1-p)^{100-k}-(100-k)p^k(1-p)^{99-k}\right) $$ $$=\sum_{k=30}^{100}\binom{100}{k}kp^{k-1}(1-p)^{100-k} - \sum_{k=30}^{99}\binom{100}{k}(100-k)p^k(1-p)^{99-k}. $$ Use $$\binom{100}{k}k =100 \binom{99}{k-1}, \quad \binom{100}{k}(100-k) =100 \binom{99}{k}$$ and rewrite $$ \frac{df(p)}{dp}=100\sum_{k=30}^{100}\binom{99}{k-1}p^{k-1}(1-p)^{100-k} - 100\sum_{k=30}^{99}\binom{99}{k}p^k(1-p)^{99-k}. $$ Denote $i=k-1$ in the first sum and rewrite $$ \frac{df(p)}{dp}=100\sum_{i=29}^{99}\binom{99}{i}p^i(1-p)^{99-i} - 100\sum_{k=30}^{99}\binom{99}{k}p^k(1-p)^{99-k} = 100 \binom{99}{29}p^{29}(1-p)^{99-29}>0. $$