put down a bs answer for #20 that doesn't simplify anything really

1 files changed, 11 insertions, 13 deletions

diff --git a/houdre/hw7.tex b/houdre/hw7.tex
index 0d4a65e..7020669 100644
--- a/houdre/hw7.tex
+++ b/houdre/hw7.tex
@@ -33,16 +33,6 @@
 
 \q2
 
-%Let the mgf of $X_i$ be $M_{X_i}(x).$
-%The sum of $n$ $X_i$ has mgf $M_{X_i}(x)^n.$
-%This has variance $(M_{X_i}(x)^n)''(0)-(M_{X_i}(x)^n)'(0)^2 =
-%(nM_{X_i}(x)^{n-1}M_{X_i}'(x))'(0)-n^2\mu^2 =
-%n(n-1)M_{X_i}(0)^{n-2}M_{X_i}'(0)^2 + nM_{X_i}(0)^{n-1}M_{X_i}''(0)
-%-n^2\mu^2 = n(n-1)\mu^2 + n\sigma^2 - n^2\mu^2 = n\sigma^2 - n\mu^2.$
-
-%The mgf of $\overline X$ is $$M_{\overline X}(x)=e^{tn^{-1}}M_{X_i}(x)^n.
-%\var\overline X = (e^{tn^{-1}}(n^{-1}M_{X_i}(x)^n+nM_{X_i}(x)^{n-1}M_{X_i}'(x)^n))' $$
-
 $$\E\left(\fr1{n-1}\sum_{i=1}^n(X_i-\overline X)^2\right)
 = \E\left(\fr1{n-1}\sum_{i=1}^n(X_i-\mu - (\overline X -
 \mu))^2\right)$$$$
@@ -125,11 +115,19 @@ $$M_{A_n}(t) = \E(e^{itn^{-1}(X_1+\cdots+X_n)}) =
 By applying the partition theorem to $M(t) = \E(e^{tX}),$ we trivially
 get $$M(t) = \sum_{k=1}^\infty \Pr(N=k)\E(e^{tX}|N=k).$$
 
+Given that $\Pr(N=k) = {1\over(e-1)k!}$ and 
 Given these specific distributions,
 $$\Pr(X\leq x) = \Pr(U_1\leq x)\cdots\Pr(U_k\leq x) = x^n \to
-f_X(x) = nx^{n-1} \to \E(e^{tX}|N=k) = \sum_{j=0}^\infty {1\over j!}t^j
-\E(n^jx^{j(n-1)}) = \sum_{j=0}^\infty {1\over
-j!}t^jn^j{1\over1+j(n-1)}.$$
+f_X(x) = nx^{n-1}$$$$ \to \E(X^j) = \int_0^1 x^jnx^{n-1}dx = \int_0^1
+nx^{n+j}/(j+n)\to \E(e^{tX}|N=k) = \sum_{j=0}^\infty {1\over
+j!}t^jnx^{n+j}/(j+n)$$$$
+\to M(t) = \sum_{k=1}^\infty {1\over(e-1)k!}\sum_{j=0}^\infty {1\over
+j!}t^jnx^{n+j}/(j+n).$$
+
 using $\E(e^{tX}) = \sum_{n=0}^\infty {t^j\over j!}\E(X^j).$
 
+The difference of $R$ and $X$ must be exponentially distributed because
+the quotient of their moment generating functions is similar to an
+exponential moment generating function.
+
 \bye