Skip to main content
fixed error in formula
Source Link

The left-hand side is the coefficient of $x^k$ in $$ \left(1+(p-1)x\right)^{r-1}\left(1+(1+p)x\right)^{r-1}=\left(1+2x+(1-p^2)x^2\right)^{r-1}\ . $$$$ \left(1+(1-p)x\right)^{r-1}\left(1+(1+p)x\right)^{r-1}=\left(1+2x+(1-p^2)x^2\right)^{r-1}\ . $$ This coefficient can be obtained via the Multinomial Theorem as $$ \sum_{a,b}\mathbf{1}\left\{ \begin{array}{c} a,b\ge 0 \\ a+b\le r-1 \\ a+2b=k \end{array} \right\}\frac{(r-1)!}{(r-1-a-b)!a!b!} 2^a (1-p^2)^b $$ where $\mathbf{1}\{\cdots\}$ is the indicator function of the conditions between the braces. Now it is immediate that the left-hand side is a decreasing function of $p$ on the interval $[0,1]$. It is thus bounded above by the value at $p=0$ which is equal to the right-hand side, by the Chu-Vandermonde identity.

The left-hand side is the coefficient of $x^k$ in $$ \left(1+(p-1)x\right)^{r-1}\left(1+(1+p)x\right)^{r-1}=\left(1+2x+(1-p^2)x^2\right)^{r-1}\ . $$ This coefficient can be obtained via the Multinomial Theorem as $$ \sum_{a,b}\mathbf{1}\left\{ \begin{array}{c} a,b\ge 0 \\ a+b\le r-1 \\ a+2b=k \end{array} \right\}\frac{(r-1)!}{(r-1-a-b)!a!b!} 2^a (1-p^2)^b $$ where $\mathbf{1}\{\cdots\}$ is the indicator function of the conditions between the braces. Now it is immediate that the left-hand side is a decreasing function of $p$ on the interval $[0,1]$. It is thus bounded above by the value at $p=0$ which is equal to the right-hand side, by the Chu-Vandermonde identity.

The left-hand side is the coefficient of $x^k$ in $$ \left(1+(1-p)x\right)^{r-1}\left(1+(1+p)x\right)^{r-1}=\left(1+2x+(1-p^2)x^2\right)^{r-1}\ . $$ This coefficient can be obtained via the Multinomial Theorem as $$ \sum_{a,b}\mathbf{1}\left\{ \begin{array}{c} a,b\ge 0 \\ a+b\le r-1 \\ a+2b=k \end{array} \right\}\frac{(r-1)!}{(r-1-a-b)!a!b!} 2^a (1-p^2)^b $$ where $\mathbf{1}\{\cdots\}$ is the indicator function of the conditions between the braces. Now it is immediate that the left-hand side is a decreasing function of $p$ on the interval $[0,1]$. It is thus bounded above by the value at $p=0$ which is equal to the right-hand side, by the Chu-Vandermonde identity.

Source Link

The left-hand side is the coefficient of $x^k$ in $$ \left(1+(p-1)x\right)^{r-1}\left(1+(1+p)x\right)^{r-1}=\left(1+2x+(1-p^2)x^2\right)^{r-1}\ . $$ This coefficient can be obtained via the Multinomial Theorem as $$ \sum_{a,b}\mathbf{1}\left\{ \begin{array}{c} a,b\ge 0 \\ a+b\le r-1 \\ a+2b=k \end{array} \right\}\frac{(r-1)!}{(r-1-a-b)!a!b!} 2^a (1-p^2)^b $$ where $\mathbf{1}\{\cdots\}$ is the indicator function of the conditions between the braces. Now it is immediate that the left-hand side is a decreasing function of $p$ on the interval $[0,1]$. It is thus bounded above by the value at $p=0$ which is equal to the right-hand side, by the Chu-Vandermonde identity.