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4

For every $p\gt1$, $r=1-1/p$ is in $(0,1)$ and, for every $x$, $(p-1)/p^{x+1}=r(1-r)^{x}$. Thus, the distribution of $X$ is geometric with parameter $r$. If $p=1$, the identities $P(X = x) = (p-1)/p^{(x + 1)}$ do not define a legitimate distribution since one would have $P(X=x)=0$ for every $x$.


3

Let $W_t$ be win today, $W_y$ be win yesterday, $L_t$ be lose today and $L_y$ be lose yesterday. We have $P(W_t)=P(W_t|W_y)\cdot P(W_y)+P(W_t|L_Y)\cdot P(L_Y)$. From the given information, this becomes $P(W_t)=.8P(W_y)+.4P(L_y)$. If the probabilities are to be stable (long-term), we should have $P(W_y)=P(W_t)$ and $P(L_y)=P(L_t)$. Thus ...


3

The main goal of the exercise is to make you realize (and use the fact) that the random variable $X$ defined as $$ X=\int_0^tY_s\mathrm ds, \qquad Y_s=W_s-\frac{s}{t}W_t, $$ is a linear combination of the gaussian family $(W_s)_{0\leqslant s\leqslant t}$ and that, as such, $X$ is itself gaussian. Hence, to fully determine the distribution of $X$, all ...


2

In general $$ \eqalign{P(X + Y \le z) &= \iint_{\{(x,y): x+y \le z\}} dx\; dy\; f_{XY}(x,y)\cr &= \int_{-\infty}^\infty dx \int_{-\infty}^{z-x} dy \; f_{XY}(x,y) }$$ If $X$ and $Y$ are independent, $f_{XY}(x,y) = f_X(x) f_Y(y)$ so this becomes $$ \int_{-\infty}^\infty dx \int_{-\infty}^{z-x} dy\;f_X(x) f_Y(y) = \int_{-\infty}^\infty dx\; ...


2

Recall that if $X \sim \mathrm{Binomial}(n,p)$, then $$\Pr[X = x] = \binom{n}{x} p^x (1-p)^{n-x}, \quad x = 0, 1, 2, \ldots, n.$$ Then use this to calculate the ratio $$\frac{\Pr[X = x+1]}{\Pr[X = x]},$$ being careful to cancel like terms.


2

From the Newton binomial expansion $$ (a+b)^{n}=\sum_{k=0}^n\binom{n}{k}a^kb^{n-k} \tag1 $$ you may deduce, by differentiating twice with respect to $a$, that $$ n(a+b)^{n-1}=\sum_{k=1}^nk\binom{n}{k}a^{k-1}b^{n-k} \tag2 $$ $$ n(n-1)(a+b)^{n-2}=\sum_{k=2}^n k(k-1)\binom{n}{k}a^{k-2}b^{n-k} \tag3 $$ Now, we have $$ ...


2

If Debra selects $N$ red cubes, then she must have selected $(3 - N)$ black cubes, so her winnings are given by: $$ W = 2N - 1(3 - N) = 3N - 3 $$ so $a = 3$ and $b = -3$.


2

I cannot say whether you will consider this as a "simplification", but you can start by writing the standard normal bivariate integral as $$\Phi_2(x,y;\rho)=\frac{1}{\sqrt {2\pi}}\int_{-\infty}^y\frac{1}{\sqrt {2\pi}\sqrt{1-\rho^2}}e^{-\frac{1}{2(1-\rho^2)}t^2}\int_{-\infty}^x e^{-\frac{1}{2(1-\rho^2)}(s^2-2st\rho)}dsdt$$ $$=\frac{1}{\sqrt ...


2

Remember this following result: Theorem A probability $P$ on $(\Omega,\mathcal{A})$ and a converging sequence of sets $A_n \in \mathcal{A}$ with limit $A \in \mathcal{A}$. Then $\lim_{n \to \infty}P(A_n) = P(A)$ In particular you have that if $A_n \to \Omega \implies P(A_n) \to 1$ and $A_n \to \emptyset \implies P(A_n) \to 0$. A CDF of a probability ...


1

Let $X=G\left(\dfrac 1 {1-\mathrm e^{-E}}\right)$. If $G$ is the inverse of $\dfrac1 {1-F}$, then the identity $X=\dfrac 1 {1-F\left(\dfrac 1 {1-\mathrm e^{-E}}\right)}$ written in the post is incorrect. Actually, $X=G\left(\dfrac 1 {1-\mathrm e^{-E}}\right)$ translates as $\dfrac1 {1-F(X)}=\dfrac 1 {1-\mathrm e^{-E}}$ hence $F(X)=\mathrm e^{-E}$ and, for ...


1

Considering $f$ and $g$ some smoothed versions of the functions $\bar f=\mathbf 1_{(0,1)}$ and $\bar g=2\mathbf 1_{(0,1/2)}$, say $f=\bar f\ast h$ and $g=\bar g\ast h$ where $h$ is a gaussian density of vanishingly small variance, one sees that the conclusion of Question 1 does not necessarily hold.


1

$$\{X\leq 1\}=\{X<\frac{1}{2}\}\cup\{\frac{1}{2}\leq X\leq 1\}$$ and these sets are disjoint so that: $$P\{X\leq 1\}=P\{X<\frac{1}{2}\}+P\{\frac{1}{2}\leq X\leq 1\}$$ or equivalently: $$P\{\frac{1}{2}\leq X\leq 1\}=P\{X\leq 1\}-P\{X<\frac{1}{2}\}$$


1

You should understand the difference between a probability density function and a cumulative distribution function. The cumulative distribution function, which in your case is $F(x)$, always gives the value for $P(X \leq x)$ So, $F(1)$ would give you $P(X\leq1)$ and $F(\frac{1}{2})$ would give you $P(X\leq\frac{1}{2})$. In order to find $P(\frac{1}{2} < ...


1

The area (probability) up to $0$ is $(1)(0.2)$, which is $0.2$. To get to area $0.5$, we need another $0.3$. So if $m$ is the median then $$\int_0^m (0.2+1.2x)\,dx=0.3.$$ The rest is calculation. We could also solve the problem geometrically. For the mode, graph the density function. It reaches its maximum at $1$.


1

If there have been four claims at most, you need to consider values of $N$ from $\color{blue}{0}$ to $4$ inclusive, hence the total probability is $$P(N=0)+P(N=1)+P(N=2)+P(N=3)+P(N=4)=\frac{5}{6}$$ For the probability of at least one outcome, the values of $N$ from $1$ to $4$ need to be considered ...


1

Well, $\mathrm{var}(X) = np(1-p)$ and $\mathrm{var}(X) = E(X²) - (E(X))^2$, so: $$E(X²) = \mathrm{var}(X) + (E(X))² = np(1-p) + n^2p^2 = np(1-p+np)$$


1

(Here, I assume that you do not know the variance of $X$; otherwise, $\mathbb{E}(X^2)$ can be easily calculated using $(1)$.) If $X \sim Bin(n,p)$, then we can write $$X = \sum_{j=1}^n Y_j$$ where $Y_j$ are independent identically distributed random variables $Y_j \sim p \delta_1+ (1-p) \delta_0$ (i.e. $\mathbb{P}(Y_j = 1) = p$, $\mathbb{P}(Y_j = 0) = ...


1

Let $\mu := p\mu_1+(1-p)\mu_2$ and $\sigma^2 := p^2\sigma_1^2 + (1-p)^2\sigma_2^2$. Then, one can show that, the pdf of $Y$ is: $$g(y)=\frac{1}{y\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(\ln(y)-\mu)^2}{2\sigma^2}\right),$$ not the one that you display (your $Y$ presentation as linear combination of log-normal variables is wrong: if $X=pX_1 +(1-p)X_2$, then ...


1

With $p$ denoting a constant in the interval $(0,1)$, suppose $X$ has density $$f(x)=p\frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right)+ (1-p)\frac{1}{\sqrt{2\pi\sigma_2^2}}\exp\left(-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right)$$ which is a mixture density. Then, $$\begin{align} F_X(\alpha) &= P\{X \leq \alpha\}\\ ...


1

$$\begin{align}\require{cancel} \Pr[X=x]&={n\choose {x}}p^{x}(1-p)^{n-x}\\ \Pr[X=x+1]&={n\choose {x+1}}p^{x+1}(1-p)^{n-(x+1)}\\ &={n\choose x+1}\left(\dfrac p{1-p}\right) p^x(1-p)^{n-x}\\ &={n\choose x+1}\left(\dfrac p{1-p}\right)\dfrac{\Pr[X=x]}{n\choose x}\\ &=\dfrac {n\choose x+1}{n\choose x}\left(\dfrac p{1-p}\right)\Pr[X=x]\\ ...


1

The sum of probabilities must still be one: $$1=\sum_{n=0}^{\infty} P(X=n)=\sum_{n=0}^{\infty} k \left(\frac9{10}\right)^n=\frac{k}{1-\frac9{10}}=10k$$ Hence $k=\frac{1}{10}$.


1

Some counterexamples to the assertion that every PDF $p$ converges to zero at $\pm\infty$ are explained in the comments, a smooth variant is a "sum-of-bumps" PDF such as$$p(x)=\sum_ng\left(\frac{x-x_n}{\sigma_n}\right),$$ where $g$ is the standard gaussian density, $\sigma_n\gt0$ for every $n$, $\sum\limits_n\sigma_n=1$, and $x_n\to\infty$. Then $p$ is ...



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