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Cauchy
Cauchy
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Assume $P_1,P_2,P_3$ different to each other pmfs. We would like to find an upper bound for $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3)$, where $D_{KL}$ is the Kullback-Leibler divergence and $*$ is convolution. By using the log sum inequality we can get $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3) \leq D_{\text{KL}}(P_1||P_2 )$. Is there any way to get a tighter bound?

Assume $P_1,P_2,P_3$ pmfs. We would like to find an upper bound for $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3)$, where $D_{KL}$ is the Kullback-Leibler divergence and $*$ is convolution. By using the log sum inequality we can get $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3) \leq D_{\text{KL}}(P_1||P_2 )$. Is there any way to get a tighter bound?

Assume $P_1,P_2,P_3$ different to each other pmfs. We would like to find an upper bound for $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3)$, where $D_{KL}$ is the Kullback-Leibler divergence and $*$ is convolution. By using the log sum inequality we can get $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3) \leq D_{\text{KL}}(P_1||P_2 )$. Is there any way to get a tighter bound?

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user44143
user44143

Assume $P_1,P_2,P_3$ pmfs. We would like to find an upper bound for $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3)$, where $D_{KL}$ is the Kullback-Leibler divergence and $*$ is convolution. By using the log sum inequality we can get $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3) \leq D_{\text{KL}}(P_1||P_2 )$. Is there any way to get a tighter bound?

Assume $P_1,P_2,P_3$ pmfs. We would like to find an upper bound for $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3)$. By using the log sum inequality we can get $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3) \leq D_{\text{KL}}(P_1||P_2 )$. Is there any way to get a tighter bound?

Assume $P_1,P_2,P_3$ pmfs. We would like to find an upper bound for $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3)$, where $D_{KL}$ is the Kullback-Leibler divergence and $*$ is convolution. By using the log sum inequality we can get $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3) \leq D_{\text{KL}}(P_1||P_2 )$. Is there any way to get a tighter bound?

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Cauchy
Cauchy
  • 153
  • 6

KL Divergence - Convolution of distributions

Assume $P_1,P_2,P_3$ pmfs. We would like to find an upper bound for $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3)$. By using the log sum inequality we can get $D_{\text{KL}}(P_1\ast P_3||P_2 \ast P_3) \leq D_{\text{KL}}(P_1||P_2 )$. Is there any way to get a tighter bound?