Let Y be a gamma-distributed random variable where ? is a

QUESTION:

Let 𝑌 be a gamma-distributed random variable where \(\alpha\) is a positive integer and \(\beta=1\). The result given in Exercise 4.99 implies that that if \(𝑦>0\),

                  \(\sum_{x=0}^{\alpha-1} \frac{y^{x} e^{-y}}{x !}=P(Y>y)\).

Suppose that \(X_{1}\) is Poisson distributed with mean \(\lambda_{1}\) and \(X_{2}\) is Poisson distributed with mean \(\lambda_{2}\), where \(\lambda_{2}>\lambda_{1}\).

a Show that \(P\left(X_{1}=0\right)>P\left(X_{2}=0\right)\).

b Let 𝑘 be any fixed positive integer. Show that \(P\left(X_{1} \leq k\right)=P\left(Y>\lambda_{1}\right)\) and \(P\left(X_{2} \leq k\right)=P\left(Y>\lambda_{2}\right)\), where 𝑌 is has a gamma distribution with \(\alpha=k+1\) and \(\beta=1\).

c Let 𝑘 be any fixed positive integer. Use the result derived in part (b) and the fact that \(\lambda_{2}>\lambda_{1}\) to show that \(P\left(X_{1} \leq k\right)>P\left(X_{2} \leq k\right)\).

d Because the result in part (c) is valid for any \(k=1\), 2, 3, . . . and part (a) is also valid, we have established that \(P\left(X_{1} \leq k\right)>P\left(X_{2} \leq k\right)\) for all \(k=0\), 1, 2, . . . . Interpret this result.

Equation Transcription:

Text Transcription:

beta=1

y>0

sum x=0 to alpha-1 y^x e^-y over x!=P(Y>y)

X_1

lambda_1

X_2

lambda_2

lambda_2>lambda_1

P(X_1=0)>P(X_2=0)

P(X_1</=k)=P(Y>lambda_1)

P(X_2</k)=P(Y>lambda_2)

alpha=k+1

beta=1

lambda_2>lambda_1

P(X_1</=k)>P(X_2</=k)

k=1

P(X_1</=k)>P(X_2</=k)