5

Almost finished. $(4y-1)^{-1}$ is a constant independent of $t$ and comes out of the limit. The limit does not exist if $\Re y>1/4$ and converges to $-1$ if $\Re y<1/4$ as $e^{-\infty}=0$, so we get $M_t(y)=(1-4y)^{-1}$ when $\Re y<1/4$. In fact the distribution you are given is an exponential distribution with parameter $\lambda=1/4$. The MGF for ...


3

Suppose such an $X$ exists. Then $Y = e^X$ is well-defined. But $$\operatorname{Var}[Y] = \operatorname{E}[Y^2] - \operatorname{E}[Y]^2 = \operatorname{E}[e^{2X}] - \operatorname{E}[e^X]^2 = 4 - 3^2 = -5,$$ so no such $Y$ exists, consequently no such $X$ exists.


2

I don't think so. Consider for now the case $\sigma^2=1$; the claim is that the only lower bound you can get is the trivial $\mathbb{E}[e^X] \geq 1$. To see why, consider any $n\geq 1$, and the random variable $X_n$ defined by $$X_n = \begin{cases} \frac{1}{n} & \text{ with probability } 1-\frac{1}{n^2+1}\\ -n & \text{ with probability } \frac{1}{n^2+...


2

Skewness is $$S = \frac{E[X^3]-3\mu\sigma^2-\mu^3}{\sigma^3}$$. In the special case of $\mu = 0$, this simplifies to $E[X^3]/\sigma^3$ Meanwhile variance is $$\sigma^2 = E[X^2]-E[X]^2$$ so if $\mu=0$ that simplifies to $E[X^2]$. So in this case, your expression $$\frac{E[X^2]}{E[X^3]}= \frac{\sigma^2}{S\sigma^3} = \frac{1}{S\sigma}$$ I don't see what the ...


2

Note that $E[X] = \mu$, hence $$ E[X^2] - \mu E[X] =E[X^2] - E[X] E[X] = E[X^2] - E^2[X] = Var(X) =\sigma^2 $$ Therefore, the skewness is \begin{align} \frac{E(X-\mu)^3}{\sigma^3} &= \frac{E[X^3] - 3\mu E[X^2] + 3\mu^2 E[X] - \mu^3}{\sigma^3}\\ & = \frac{E[X^3] - 3\mu ( E[X^2] - \mu E[X]) - \mu^3}{\sigma^3}\\ & = \frac{E[X^3] - 3\mu \sigma^2 - \...


1

If $W \sim \mathcal{N}(0, a)$, its density function is given by $$f_{W}(w) = \dfrac{1}{\sqrt{a} \cdot \sqrt{2\pi}} \cdot e^{-\frac{w^2}{2a}}$$ for $w \in \mathbb{R}$. Then, use the fact that $\sqrt{a}\cdot \sqrt{2\pi} = \sqrt{2a\pi}$.


1

For the sum of five dice, a roll of $k$ is just as likely as a roll of $35-k$, for any $k\in \mathbb Z$. Therefore, \begin{aligned} P(\text{white $-$ red}=0) &=\sum_{k=5}^{30} P(\text{white }=k)P(\text{red }=k) \\&=\sum_{k=5}^{30} P(\text{white }=k)P(\text{red }=35-k) \\&=P(\text{white + red }=35) \end{aligned} So you just need to find the ...


1

So for the first part we can write, \begin{align} M_{S_N}(t) = \mathbb{E}[e^{tS_{N}}] &= \mathbb{E}[e^{t\sum_{i=1}^{N}X_i}] \\ &= \mathbb{E}\left[\mathbb{E}[e^{t\sum_{i=1}^{N}X_i}|N]\right], \hspace{5mm} \text{l.i.e} \\ &= \mathbb{E}\left[(\mathbb{E}[e^{tX_1}])^N\right] \\ &= \mathbb{E}\left[e^{\ln(M_X(t))N}\right] \\ &= M_N(\ln(M_X(t))). ...


1

If you want to avoid vector/matrix stuff, just write $$M_X(t) = \exp\left(\sum_i t_i \mu_i + \frac{1}{2}\sum_i \sum_j\Sigma_{ij} t_i t_j\right)$$ and find the partial derivatives using basic calculus. Doing this can actually give you insight into the "slick" ways of doing this using vector/matrix differentiation rules. If you want to use shortcuts:...


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