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Iosif Pinelis
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First here, $E(f(X(t)|X(0)=x)$ is, not a conditional probability, but a conditional expectation. Second, the event $X(0)=x$ is certain to occur; therefore, conditioning on it does not affect the expectations or probabilities. So, $$E(f(X(t)|X(0)=x)=Ef(X(t).$$ Finally, $$Ef(X(t))=\sum_{k=0}^\infty\sum_{l=0}^\infty f\Big(x+\frac{k-l}{\sqrt n}\Big)\frac{(t\lambda)^{k+l}}{k!\,l!}e^{-2\lambda}.$$$$Ef(X(t))=\sum_{k=0}^\infty\sum_{l=0}^\infty f\Big(x+\frac{k-l}{\sqrt n}\Big)\frac{(t\lambda)^{k+l}}{k!\,l!}e^{-2t\lambda}.$$

First here, $E(f(X(t)|X(0)=x)$ is, not a conditional probability, but a conditional expectation. Second, the event $X(0)=x$ is certain to occur; therefore, conditioning on it does not affect the expectations or probabilities. So, $$E(f(X(t)|X(0)=x)=Ef(X(t).$$ Finally, $$Ef(X(t))=\sum_{k=0}^\infty\sum_{l=0}^\infty f\Big(x+\frac{k-l}{\sqrt n}\Big)\frac{(t\lambda)^{k+l}}{k!\,l!}e^{-2\lambda}.$$

First here, $E(f(X(t)|X(0)=x)$ is, not a conditional probability, but a conditional expectation. Second, the event $X(0)=x$ is certain to occur; therefore, conditioning on it does not affect the expectations or probabilities. So, $$E(f(X(t)|X(0)=x)=Ef(X(t).$$ Finally, $$Ef(X(t))=\sum_{k=0}^\infty\sum_{l=0}^\infty f\Big(x+\frac{k-l}{\sqrt n}\Big)\frac{(t\lambda)^{k+l}}{k!\,l!}e^{-2t\lambda}.$$

Source Link
Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229

First here, $E(f(X(t)|X(0)=x)$ is, not a conditional probability, but a conditional expectation. Second, the event $X(0)=x$ is certain to occur; therefore, conditioning on it does not affect the expectations or probabilities. So, $$E(f(X(t)|X(0)=x)=Ef(X(t).$$ Finally, $$Ef(X(t))=\sum_{k=0}^\infty\sum_{l=0}^\infty f\Big(x+\frac{k-l}{\sqrt n}\Big)\frac{(t\lambda)^{k+l}}{k!\,l!}e^{-2\lambda}.$$