3

Yes, provided that $e^{X}$ is integrable. It is because $\sigma(Y)=\sigma(e^{Y})$. Proof: Clearly $e^{Y}$ is $\sigma(Y)$-measurable, so $\sigma(e^{Y})\subseteq\sigma(Y)$. On the other hand, $Y=\ln\left(e^{Y}\right)$ which is $\sigma(e^{Y})$-measurable, so $\sigma(Y)\subseteq\sigma(e^{Y})$.


2

Guide: Note that $$M_X(t)=E[e^{Xt}]=\sum P(X=x)e^{xt}$$ Hence I can read of from the first term of MGF that $P(X=-2)=\frac16$. Try to read off the other terms and you should be able to answer the question.


2

Sum of two moment generating functions can never be a moment generating function. This is because any moment generating function had the value $1$ at $0$.


2

Without any further rearrangement, if you plug in $t=0$ at the left hand side of your first equation you will see that such a function cannot exist.


1

To justify the interchange of limit and integral, estimate the difference quotient as follows: Let $|t|<\delta<1$, $$ \left|\frac{e^{-t(\mu-x)}-1}{t}\right|e^{-x}\leq e^{-x}\sup_{|t|<\delta}|x-\mu|e^{-t(\mu-x)}\leq e^{-(x-\delta|x-\mu|)}|x-\mu| $$ Which is integrable. Now apply the DCT to find $$ \lim_{t\to 0}\int_0^\infty \frac{e^{-t(\mu-x)}-1}{...


1

I don't understand why one would use MGF for this . You get it from definition: $P(Y<c)=P(X<\frac {c-a} {b-a}) =\frac {c-a} {b-a}$ for $c$ between $a$ and $b$ and this is the definition of uniform distribution.


1

From the question "Also, what will be the limits? Will it be from x to 1 or from 0 to 1?" I will assume that the proper joint density is $$ f_{X,Y}(x,y) = 8xy\cdot\mathsf 1_{0<x<y<1}. $$ To find the moment-generating function of $Y$, we first need to determine its marginal distribution. To do this, we integrate over all possible values of $X$: $$ ...


1

For $y\geqslant 0$ we have $1\geqslant e^{-y}=\sum_{k=0}^{\infty}(-1)^k y^k/k!$; integrating this $m$ times, we get $$\frac{y^m}{m!}\geqslant\sum_{k=0}^{\infty}\frac{(-1)^k y^{k+m}}{(k+m)!}\underset{k+m=j}{=}(-1)^m\sum_{j=m}^{\infty}(-1)^j\frac{y^j}{j!}.$$


1

Now you have to apply the formula for the partial sum of a geometric series (given hint): $$\sum_{k=1}^n r^k=r\cdot \frac{r^n-1}{r-1}$$ For $|r|<1$ the series converges: $\sum\limits_{k=1}^{\infty} r^k=\lim\limits_{n \to \infty }r\cdot \frac{r^n-1}{r-1}=r\cdot \frac{0-1}{r-1}=\frac{r}{1-r} \qquad (*)$ Next we we simplify the sum: $$M_{X}(t) = p\sum_{...


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