STEP II 2013 12

(i)

Note that:

\begin{align*}\mathrm E(X) &= 1 \times \mathrm P(U = 1) + 3 \times \mathrm P(U = 3) + 5 \times \mathrm P(U = 5) + \ldots \\ & = 1 \times e^{-\lambda} \lambda + 3 \times e^{-\lambda} \frac {\lambda^3} {3!} + 5 \times e^{-\lambda} \frac {\lambda^5} {5!} + \ldots \\ & = e^{-\lambda} \lambda \left(1 + \frac {\lambda^2} {2!} + \frac {\lambda^2} {4!} + \ldots\right) \\ & = e^{-\lambda} \lambda \alpha\end{align*}

Similarly:

\begin{align*}\mathrm E(Y) & = 2 \times \mathrm P(U = 2) + 4 \times \mathrm P(U = 4) + \ldots \\ & = 2 \times e^{-\lambda} \frac {\lambda^2} {2!} + 4 \times e^{-\lambda} \frac {\lambda^4} {4!} \\ & = e^{-\lambda} \lambda \left(\lambda + \frac {\lambda^3} {3!} + \ldots\right) \\ & = e^{-\lambda} \lambda \beta\end{align*}

(ii)

Before we proceed, note that \alpha + \beta = e^\lambda, so that e^{-\lambda} = \dfrac 1 {\alpha + \beta}. With that we can write:

\displaystyle \mathrm E(X) = \frac {\lambda \alpha} {\alpha + \beta}

and:

\displaystyle \mathrm E(Y) = \frac {\lambda \beta} {\alpha + \beta}

We have:

\begin{align*} \mathrm E(X^2) & = 1 \times \mathrm P(X = 1) + 3^2 \times \mathrm P(X = 3) + 5^2 \times \mathrm P(X = 5) + \ldots \\ & = e^{-\lambda} \left(\frac {\lambda^1} {1!} + \frac {3 \lambda^3} {2!} + \frac {5 \lambda^5} {4!} + \ldots\right) \\ & = e^{-\lambda} \lambda \left(1 + \frac {3 \lambda^2} {2!} + \frac {5 \lambda^4} {4!} + \ldots\right)\end{align*}

We can write:

\begin{align*}1 + \frac {3 \lambda^2} {2!} + \frac {5 \lambda^4} {4!} + \ldots &= 1 + \frac {(2 + 1) \lambda^2} {2!} + \frac {(4 + 1) \lambda^4} {4!} + \ldots \\ & = 1 + \lambda^2 + \frac {\lambda^2} {2!} + \frac {\lambda^4} {3!} + \frac {\lambda^4} {4!} + \ldots + \\ & = \left(1 + \frac {\lambda^2} {2!} + \frac {\lambda^4} {4!} + \ldots\right) + \lambda \left(\lambda + \frac {\lambda^3} {3!} + \frac {\lambda^5} {5!} + \ldots\right) \\ & = \alpha +\lambda \beta\end{align*}

We have:

\displaystyle \mathrm E(X^2) = \frac {\lambda} {\alpha + \beta} \left(\alpha + \lambda \beta\right) = \frac {\lambda \alpha + \lambda^2 \beta} {\alpha + \beta}

So that:

\displaystyle \mathrm{var}(X) = \frac {\lambda \alpha + \lambda^2 \beta} {\alpha + \beta} - \frac {\lambda^2 \alpha^2} {(\alpha + \beta)^2}

as required.

We also have:

\begin{align*}e^\lambda \mathrm E(Y^2) &= 2^2 \frac {\lambda^2} {2!} + 4^2 \frac {\lambda^4} {4!} + 6^2 \frac {\lambda^6} {6!} + \ldots \\ & = 2 \lambda^2 + \frac {4 \lambda^4} {3!} + \frac {6 \lambda^6} {5!} + \ldots \\ & = \lambda^2 + \lambda^2 + \frac {(3 + 1) \lambda^4} {3!} + \frac {(5 + 1) \lambda^6} {5!} + \ldots \\ & = \lambda^2 + \lambda^2 + \frac {\lambda^4} {2!} + \frac {\lambda^4} {3!} + \frac {\lambda^6} {4!} + \frac {\lambda^6} {5!} + \ldots \\ & = \lambda \left(\lambda + \frac {\lambda^3} {3!} + \frac {\lambda^5} {5!} + \ldots\right) + \lambda^2 \left(1 + \frac {\lambda^2} {2!} + \frac {\lambda^4} {4!} + \ldots\right) \\ & = \lambda \beta + \lambda^2 \alpha\end{align*}

So that:

\displaystyle \mathrm E(Y^2) = \frac {\lambda \beta +\lambda^2 \alpha} {\alpha + \beta}

So:

\displaystyle \mathrm{var}(Y) = \frac {\lambda \beta + \lambda^2 \alpha} {\alpha + \beta} - \frac {\lambda^2 \beta^2} {(\alpha + \beta)^2}

Note that:

\mathrm{var}(X + Y) = \mathrm{var}(U) = \lambda

and:

\displaystyle \mathrm{var}(X) + \mathrm{var}(Y) = \frac {\lambda(\alpha + \beta) + \lambda^2 (\alpha + \beta)} {\alpha + \beta} - \frac {\lambda^2 (\alpha^2 + \beta^2)} {(\alpha + \beta)^2} = \lambda

So that:

\displaystyle \lambda + \lambda^2 - \frac {\lambda^2 (\alpha^2 + \beta^2)} {(\alpha + \beta)^2} = \lambda

Or:

\displaystyle \lambda^2 \left(1 - \frac {\alpha^2 + \beta^2} {(\alpha + \beta)^2}\right) = 0

which we can write as:

\displaystyle \lambda^2 \frac {2 \alpha \beta} {(\alpha + \beta)^2} = 0

since \displaystyle \frac {2 \alpha \beta} {(\alpha + \beta)^2} > 0, we must have \lambda = 0. So no, there are no non-zero values of \lambda for which \mathrm{var}(X) + \mathrm{var}(Y) = \mathrm{var}(X + Y).

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