The mean of the
distribution is λ and the mean of the
distribution is 1/λ.

Question

The mean of the 
 distribution is λ and the mean of the 
 distribution is 1/λ.

BlackTom AI · Accepted Answer

For a Poisson distribution with parameter λ, the expected value equals λ, and for an exponential distribution with rate parameter λ, the expected value equals 1/λ.

Relevant concepts:
- Poisson(λ) is a discrete distribution on nonnegative integers with probability mass function P(X=k)=e^{-λ} λ^k / k!, k=0,1,2,...; its mean E[X] can be computed by summing k·P(X=k).
- Exponential(λ) is a continuous distribution on x≥0 with density f(x)=λ e^{-λ x}; its mean E[X] is the integral ∫_0^∞ x f(x) dx, computed by integration by parts.

Step-by-step reasoning and calculations:

1) Poisson mean:
E[X] = Σ_{k=0}^∞ k·e^{-λ} λ^k / k!
Separate k=0 term (zero) and re-index:
E[X] = e^{-λ} Σ_{k=1}^∞ k λ^k / k!
Write k/k! = 1/(k-1)!, so
E[X] = e^{-λ} Σ_{k=1}^∞ λ^k /(k-1)! = e^{-λ} λ Σ_{k=1}^∞ λ^{k-1}/(k-1)!
Let j = k-1 (j from 0 to ∞):
E[X] = e^{-λ} λ Σ_{j=0}^∞ λ^j / j! = λ e^{-λ} e^{λ} = λ.
Thus the Poisson mean is λ.

2) Exponential mean:
E[X] = ∫_0^∞ x λ e^{-λ x} dx.
Use integration by parts: let u = x, dv = λ e^{-λ x} dx ⇒ du = dx, v = -e^{-λ x}.
Then E[X] = [ -x e^{-λ x} ]_0^∞ + ∫_0^∞ e^{-λ x} dx.
The boundary term is 0 (since x e^{-λ x} → 0 as x→∞ and is 0 at 0). The remaining integral is ∫_0^∞ e^{-λ x} dx = (1/λ).
So E[X] = 1/λ.

Why other distributions would be incorrect:
- Normal, uniform, binomial, etc., have means that depend on different parameters (mean μ for Normal; np for Binomial), so they do not match the stated λ and 1/λ relationships.
- The Poisson is discrete (counts) and the exponential is continuous (waiting time), matching the typical mean forms shown.

Conclusion: The blanks should be filled with Poisson and exponential respectively, which matches the provided answer ["Poisson", "exponential"].

The mean of the distribution is λ and the mean of the distribution is 1/λ.未知题型

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