Expected Value of 2^X and 2^-X for Geometric and Poisson Distributions?

In summary, for the distributions of $$X\sim Geom(p)$$ and $$X\sim Pois(\lambda)$$, the expectations of $$E[2^X]$$ and $$E[2^{-X}]$$ are finite for all values of the parameter. However, the expected value for both cases seems to be divergent and cannot be calculated using the given formulas. A possible approach could be using a transform to simplify the calculations, such as $$P[2^X>a]=P[e^{Xln2}>a]$$ when finding the distribution of $$E[2^X]$$.
  • #1
Cedric Chia
22
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TL;DR Summary
Help find E[2^X] and E[2^-X] if X~Geom(p) and if X~Pois(lambda)
For the following distributions find $$E[2^X]$$ and $$E[2^{-X}]$$ if finite. In each case,clearly state for what values of the parameter the expectation is finite.

(a) $$X\sim Geom(p)$$
(b) $$X\sim Pois(\lambda)$$

My attempt:

Using LOTUS and $$E[X]=\sum_{k=0}^{\infty}kP(X=k)=\frac{1-p}{p}$$
thus
$$E[2^X]= \frac{1-p}{p} \sum_{k=0}^{\infty}\frac{2^k}{k}$$
and
$$E[2^{-X}]= \frac{1-p}{p} \sum_{k=0}^{\infty}\frac{1}{k2^k}$$
for $$X\sim Geom(p)$$
but the expected value in either case seems to be divergent, I don't know how to continue, please help.
 
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  • #2
Cedric Chia said:
$$E[2^X]= \frac{1-p}{p} \sum_{k=0}^{\infty}\frac{2^k}{k}$$
I don't think this step is correct. I suggest that you start from ##E[f(X)] = \sum_k {f(k)p(k)}##.
 
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  • #3
Along what tnich said, maybe you can use a transform like:

##P[2^X >a]=P[ e^{Xln2}>a]=...##

A simpler example: Given X, if you want to find the distribution of ##X^2## , we use:

##P[X^2>a]= P[ X > a^{1/2}] ##
 

FAQ: Expected Value of 2^X and 2^-X for Geometric and Poisson Distributions?

What is the expected value of 2^X?

The expected value of 2^X is the average value that would be obtained if the experiment of raising 2 to a power X is repeated a large number of times. It is calculated by multiplying each possible outcome by its probability and then summing all of these values.

What is the expected value of 2^-X?

The expected value of 2^-X is the average value that would be obtained if the experiment of raising 2 to the negative power of X is repeated a large number of times. It is calculated by multiplying each possible outcome by its probability and then summing all of these values.

How is the expected value of 2^X calculated?

The expected value of 2^X is calculated by multiplying each possible outcome by its probability and then summing all of these values. This can be represented mathematically as E(2^X) = ∑ (x * P(X=x)), where x is each possible outcome and P(X=x) is the probability of that outcome.

What factors affect the expected value of 2^X?

The expected value of 2^X is affected by the range of possible outcomes and their corresponding probabilities. If the range of outcomes is larger, the expected value will be higher. Additionally, the probabilities assigned to each outcome will also impact the expected value.

How is the expected value of 2^-X different from the expected value of 2^X?

The expected value of 2^-X is different from the expected value of 2^X because the range of possible outcomes and their corresponding probabilities are different. In the case of 2^-X, the outcomes are all fractions or decimals, whereas in 2^X, the outcomes are all positive integers. This results in a different calculation and expected value for each scenario.

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