Hand in: Refer to Example C pg 117 (Rice). Show that E=0 xp(x) = 1 if X~Poi(2). Expand the steps.

Hand in: Refer to Example C pg 117 (Rice). Show that E=0 xp(x) = 1 if X~Poi(2). Expand<br>the steps.<br>

Extracted text: Hand in: Refer to Example C pg 117 (Rice). Show that E=0 xp(x) = 1 if X~Poi(2). Expand the steps.
EXAMPLEC Poisson Distribution The expected value of a Poisson random variable is k入 E(X) = ) k! 入ペー! = le) (k 1)! Σ j! = de- Since (A/j!) = e, we have E(X) = À. The parameter À of the Poisson distribution can thus be interpreted as the average count.

EXAMPLEC<br>Poisson Distribution<br>The expected value of a Poisson random variable is<br>k入<br>E(X) = )<br>k!<br>入ペー!<br>= le)<br>(k 1)!<br>Σ<br>j!<br>= de-<br>Since (A/j!) = e, we have E(X) = À. The parameter À of the Poisson<br>distribution can thus be interpreted as the average count.<br>

Extracted text: EXAMPLEC Poisson Distribution The expected value of a Poisson random variable is k入 E(X) = ) k! 入ペー! = le) (k 1)! Σ j! = de- Since (A/j!) = e, we have E(X) = À. The parameter À of the Poisson distribution can thus be interpreted as the average count.

Jun 05, 2022

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Hand in: Refer to Example C pg 117 (Rice). Show that E=0 xp(x) = 1 if X~Poi(2). Expand the steps.

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