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1

The probability to get at least a four in one die is $1/2$. Therefore, the probability to get three or less in $N$ throws is $$\frac1{2^N}$$ and the probability to get a four or more in exactly one throw is $$\frac N{2^N}$$ The $N$ comes from the fact that the "good throw" can be the first, the second, ..., or the $N$-th. Thus, the probability to get four ...


0

You have forgotten minus sign from $P(X+Y=n) = \frac{(\lambda + \gamma)^n e^{\lambda + \gamma}}{n!}$ It is supposed to be $P(X+Y=n) = \frac{(\lambda + \gamma)^n e^{-\lambda - \gamma}}{n!}$ Multiply sum with $e^{2(\lambda + \gamma)}$ and sum adds up to one.


0

Your error is that the correct PMF for $X+Y$ is $$\Pr[X + Y = n] = e^{-(\lambda + \gamma)} \frac{(\lambda+\gamma)^n}{n!},$$ whereas in your solution you are missing the negative sign. The resulting correction will cause the exponential terms to cancel. The result should be a binomial distribution for a suitable choice of parameters.


1

Your answers are correct. Perhaps tell us why you thought they were incorrect, because it's important to improve on the intuition that led you to think this.


1

Hint: Fill up details in the following: With $\;X\sim B(n,p)\;$ and using the identity $\;i\binom ni=n\binom{n-1}{i-1}\;$: $$E(X^k)=\sum_{i=0}^n i^k\binom ni p^i(1-p)^{n-i}=\sum_{\color{red}{i=1}}^n i^k\binom ni p^i(1-p)^{n-i}=$$ $$=np\sum_{i=1}^n i^{k-1}\binom{n-1} {i-1} p^{i-1}(1-p)^{n-i}\stackrel{m:=i-1}=np\sum_{m=0}^n (m+1)^{k-1}\binom{n-1} m ...


0

If $\operatorname{ess\,sup}(X)\gt0$, set $0\lt\lambda\lt\operatorname{ess\,sup}(X)$, $t\gt0$, and $0\lt\epsilon\lt\lambda$. Then $$ \begin{align} \int_{x\gt\lambda-\epsilon}xe^{tx}\,\mathrm{d}\mu(x) &\ge\int_{x\gt\lambda}xe^{tx}\,\mathrm{d}\mu(x)\\ &\ge\mu\{x:x\gt\lambda\}\lambda e^{t\lambda}\tag{1} \end{align} $$ and $$ \begin{align} \int_{0\le ...


0

The approach through the cumulative distribution function is reasonable, but there are troubles with the algebra. I am interpreting $\log$ as the natural logarithm. Minor modification will take care of things if we interpret $\log$ as logarithm to the base $10$. We have for suitable $y$ $$\Pr(Y\le y)=\Pr\left(-\frac{\log(1-X)}{\lambda}\le ...


0

Hmm. Say $X$ is uniformly distributed on $[-2,1]$. I'd say the essential supremuum of $X$ was $2$; from the way the question's turning out I gather you'd say it was $1$? Assume first that $X$ is essentially bounded above. Now $p'(t)=\mathbb E[Xe^{tX}]/\mathbb E[e^{tX}]$. Say the essential sup of your $X$ is $S$. Then $X\le S$ almost surely; since ...


1

Integration by parts is the right idea, but let's start with the left side of $\displaystyle\int_{0}^{\infty}(1-F(x))\,dx - \int_{-\infty}^{0}F(x)\,dx = E(X)$ instead of the right side. We have $\displaystyle\int_{0}^{\infty}(1-F(x))\,dx = \underbrace{\left[x(1-F(x))\right]_{0}^{\infty}}_{0}-\int_{0}^{\infty}-xF'(x)\,dx = \int_{0}^{\infty}xf(x)\,dx$, and ...


0

Let at time t, the first player has moved a distance R1 while second player is situated at a point in circumference and has caught the first player at the same time t. Let the distance R2 be between his initial point in the circumference to the point where he caught the first player. Let R be the radius of the circular field. The first player randomly ...


1

When someone says that a baby is born every (instert a period of time), they actually mean that this is true on average. It can happen in a small village that three babies are born on the same day. Many of these deviations from the average get washed out if you have enough data. But some things don't. Several things affect when people are born. A mother's ...


3

The cited statistic, "every second 3 babies are born" is just a rough average. In truth, human birth rates are extremely seasonal (the weather having a dramatic effect on human behavior). Here is a reference: http://www.sas.upenn.edu/~valeggia/pdf%20papers/birth_seasonality.pdf


0

Having $n$ different machines all break down can be interpreted as the same machine breaking down $n$ times, since the the time to breaking is exponentially distributed and thus memoryless. Let $X \sim \operatorname{Poisson}(\lambda)$ be the random variable that models how many times the machine breaks down in one time step. We have $\lambda = ...


0

0.5 probability of player 2 winning means that if player 1 runs perpendicular to the line joining the initial positions of the players, then player 2 will catch him on the boundary of the circle. In that case the path of player 2 will be the hypotenuse of an isosceles right with side length $r$, the radius of the circle. the length of this path is $l=\sqrt ...


2

$X,Y,Z\sim_{iid}\text{Exp}(1)$ because $F_{X,Y,Z}(x,y,z)=F_X(x)F_Y(y)F_Z(z)$ for all $(x,y,z)\in\mathbb{R}^3$. Denote $S_n=\sum_{i=1}^nX_i$ where $X_i\sim_{iid}\text{Exp}(\lambda)$. Distribution of $S_n/n$ follows $\text{Gamma}(n,n\lambda)$ because ...


1

Consider $F(a) = \Bbb{P}(\frac{X+Y + Z}{3} \leq a) =\Bbb{P}(X+Y + Z \leq 3a)$. Now calculate $$F(a) = \int_0^{3a} \int_0^{3a-x} \int_0^{3a - x - y} e^{-z}e^{-y}e^{-x} \, dz\,dy\,dx = \\ = \int_0^{3a} \int_0^{3a-x} (1 - e^{-3a + x + y})e^{-y}e^{-x} \,dy\,dx \\ = \int_0^{3a} \int_0^{3a-x} (e^{-y} - e^{-3a + x} )e^{-x} \,dy\,dx \\ = \int_0^{3a} (1 - ...


0

$$P(X > 3) = \frac{1}{2}\left(1 + \mathrm{erf}\left(\frac{\mu - 3}{\sigma \sqrt{2}}\right)\right),$$ so $\mu = 3 + \sigma \sqrt{2}\, \mathrm{erf}^{-1}(2 c - 1)$, where $c = 0.8413$. $(X + Y)/2$ has normal distribution with $\mathsf{E} (X+Y)/2 = \mu$ and $\mathrm{Var}((X + Y)/2) = \sigma^2 / 2$. Therefore, $P((X + Y)/2 > 3) = \frac{1}{2} (1 + ...


1

Let we assume that $u(y)$ is the density of $X$ and $$ f_\tau(y) = \frac{1}{\tau\sqrt{2\pi}}e^{-\frac{y^2}{2\tau^2}} $$ is the density of $Z_\tau$. Then $g_\tau = u * f_\tau$, and since $f_\tau$ is differentiable, $g_\tau$ is differentiable and: $$ g_\tau' = u * f_t'.$$ So we have: $$ g_\tau(y) = \int_{\mathbb{R}} ...


2

Let us assume that $\Omega$ is finite or countable. I think the main confusion here is the relationship between a probability measure and a random variable. A probability measure $\mathbb P$ is a function $\mathbb P:\mathcal F \rightarrow [0, 1]$ for some $\sigma$-algebra. A random variable, on the other hand, is a function $X: \Omega \rightarrow S$ where ...


2

$Z=(X,Y)'$ is not necessary bivariate normal (it works if $X$ and $Y$ are independent). There are lots of counterexamples (e.g. link). Another example: $\psi_1$ and $\psi_2$ are independent standard normal r.v.s and $$(X, Y)=(\psi_1,|\psi_2|)1\{\psi_1\ge 0\}+(\psi_1,-|\psi_2|)1\{\psi_1< 0\}$$ which is not bivariate normal. However, marginals $X$ and $Y$ ...


0

There are two ways you could tackle the problem: (a) Use "change of variables" from $X$ and $V$ to $X$ and $Y$ and then find the required conditional distribution and expectation. (b) Use the "direct" method, which is what you've done. For this particular problem, I think (a) is a slightly simpler but I will use (b) here since it is in line with your ...


0

It is true that the ,025 and .975 quantiles of $Binom(n = 903, p = 0.03)$ are 17 and 38, respectively. However, your conclusion about significance does not make sense because you have not stated null and alternative hypotheses. (If you were testing $H_0: p = 0.03$ against $H_A: p \ne 0.03$, then getting 17 successes in 903 trials would be borderline ...


1

Hint: start with small $n$, e.g. $n=2$ or $n=3$. When $n=2$, we've got: $\mathbb{P}[X=1]=0$ (can't have a repeat chip when there are no prior chips) $\mathbb{P}[X=2]=\frac{1}{2}$ (The chance of redrawing the chip taken on turn 1 is $\frac{1}{2}$--there's only one other chip. $\mathbb{P}[X=3]=\frac{1}{2}$ (If $X!=2$, then in the first two turns we took ...


0

The probability that someone can draw $X$ coins from the bag without drawing a coin which has already been drawn is given by the recurrence relation $$\mathbb{P}(1)=\frac{1}{n};$$ $$\mathbb{P}(X)=\frac{n-X+1}{n}\mathbb{P}(X-1).$$ The reasoning here is that, on the $X^{\text{th}}$ draw, there will be $n-X$ coins which we will not have drawn yet out of $n$ ...


0

$f(b)$ can actually be defined in an entirely arbitrary way since its value has no effect on the distribution function $F$. Essentially, density functions are not unique if we only require that $\int_A f(x) \, dx = P(X \in A)$. Defining it as a derivative is convenient because then this automatically holds over the support of $X$, but oftentimes this ...


1

Since $X$ has a symmetic distribution about $0$, you have $P(X-0 \le x) = P(0-X \le x)$, i.e. $P(X \le x) = P(-X \le x)$. Now consider the cumulative distribution function $P(Z \le x)$. This is $P(XY \le x)=P(XY \le x|Y=1)P(Y=1)+P(XY \le x|Y=-1)P(Y=-1)$ which is $P(X \le x|Y=1)P(Y=1)+P(-X \le x|Y=-1)P(Y=-1)$ which by the symmetry and independence is ...


2

If $X$ and $Y$ are independent and $Y$ takes values in $\{-1,1\}$ while the distribution of $X$ is symmetric about $0$, then $XY$ has the same distribution as $X$. Hint: condition on $Y$.


0

For the transition matrix, we require that P be row-stochastic, so for all i, $\sum\limits_{j}p_{ij} = 1$. In particular, for $i=0$, since $\sum\limits_{j=0}^{\infty}{(\frac{5}{6})^j} = \dfrac{1}{1 - \frac{5}{6}} = 6$ only the first definition (with the exponent of $j$ not $j-1$) makes sense. The exponent $j-1$ would only make sense if the middle ...


0

Choose the probability space $([0,1],\mathscr{B},m)$. ($\mathscr{B}$ consists of all Borel sets of $[0,1]$, $m$ is the Lebesgue measure.) Let $X_{2n}(\omega)=\omega$, $X_{2n-1}(\omega)=1-\omega$. Show that $X_{2n}$ and $X_{2n-1}$ have the same distribution. (Uniform distribution.) Show that $\{X_n\}$ does not converge in probability.


1

The increments are stationary. Since $B_t - B_s$ is the increment over the interval $[s, t]$, it is the same in distribution as the incremeent over the interval $[s-s, t-s] = [0,t-s]$. Hence, $$B_t-B_s \sim B_{t-s}-B_0.$$ But $B_0 = 0$ almost surely, so that: $$B_t-B_s \sim B_{t-s}.$$ Finally, $B_{t-s} \sim \mathcal{N} (0,t-s)$. We didn't use the ...


0

There are no hard and fast rules. In general, the key to modelling is identifying the key aspects of the question you are interested in, determine if they are quantifiable in a meaningful way, then map those quantities to some model you develop. Overall, the key is the ability to abstract the quantitative elements of a problem and associate them with ...


1

$$G_{S_n}(z)=\left[G_{X_{n1}} (z)\right]^n=\left[\mathbb{E}z^{X_{n1}}\right]^n=\left[\frac{z}{n}+\left(\frac{z}{n}\right)^2+\left(1-\frac{1}{n}-\frac{1}{n^2}\right)\right]^n$$ $$\lim_{n\rightarrow \infty}G_{S_n}(z)=e^{z-1}$$ which corresponds to the probability generating function of $\text{Poisson(1)}$. Note: For large $n$ ...


1

One approach : First consider all of the possible orders in which component $1$ can be the second component to fail. Event $A$: $2$ fails, then $1$, then another component Event $B$: $3$ fails, then $1$, then another component Event $C$: ... (continuation) Event $D$: $n$ fails first, then $1$, then another component Then computing the probability for ...


0

An elementary example can be found by using a uniformly-distributed random variable with positive probability on [0,1] versus the uniform distribution with support (0,1). Both RVs have the same CDF (theyre equal at every point), but different densities (as a result of the behaviour at the endpoints).


0

The probability density function is defined for a specific measure $\mu$ on $\mathbb{R}$ (usually, the Lebesgue measure), and the integrals $\int_\mathbb{R} f$ and $\int_{I}f$ should be written as $\int_\mathbb{R} f d \mu$ and $\int_I f d\mu$. The probability density function is defined $\mu$-almost everywhere uniquely (see ...


0

As a first step, use the fundamental counting principle, which tells you that {number of different coffee drinks}= {number of sizes} * {number of varieties} * {number of flavor combinations}. The tricky part of the question is to count the number of flavor combinations. A customer can choose to add 0, 1, . . . , or 4 syrup flavors. To determine the number ...


0

I think you meant for the expression to be $\text{Var}(\sum_{i=1}^{N} L_i)$ where $L_i$ is the loss of the $i^\text{th}$ hurricane and $N$ is the number of hurricanes. Using the total variance formula this would be decomposed as (assuming hurricane losses are i.i.d.), $$ \begin{align} \text{Var}\left (\sum_{i=1}^{N} L_i \right ) &= \text{Var}\left [ ...


1

"In the end, only the expectation of Y is needed, not its distribution." This a typical famous university exam question. As Did already mentioned $E(X^n)=0$ for n being odd. Now for $n$ even ($2m$ say and WLOG assume $\sigma^2=1$), $$E(X^{2m})=2\int_0^{\infty} \frac{x^{2m}\exp(-x^{2}/2)}{\sqrt{2\pi}}dx$$ Substitute, $y=x^2/2$ to make the above as a $Gamma$ ...


0

Given that the data are discrete, the conditional distribution will also be discrete. In that case, the shifted exponential doesn't make much sense to me. Consider a left censored negative binomial distribution on $T_i$. If I understand correctly, $T_i = T_i^* \iff T_i^* > \delta$. In that case the distribution should be $$P(T_i = t) = \begin{cases} ...


0

If $P(X=1)+P(X=2)+P(X=4)=1$ then $P(x=i)=0$, unless $i=1$ or $2$ or $4$, that is, the pmf for the event $\{X=3\}$ does exist: $P(X=3)=0$. The cdf, by one of the possible two definitions, is $$F_1(x)=P(X<x)=\begin{cases} 0,&\text{ if } x\le 1\\ P(X=1),&\text{ if } 1<x\le 2\\ P(X=1)+P(X=2),&\text{ if } 2<x\le 4\\ 1,&\text{ if } ...


0

The Gamma distribution is commonly used. Your plot does look like it might have a sharper peak than that, in which case you might use an inverse Gaussian distribution. Both these distributions have two parameters so you'd need to fit them to your data using Maximum Likelihood Estimation. The equations for these are given in the Wikipedia pages.


2

A closed form expression is provided in the following paper. SV Amari, RB Misra, Closed-form expressions for distribution of sum of exponential random variables, IEEE Transactions on Reliability, 46 (4), 519-522.


0

The most basic difference between probability mass func and probability density function is that pmf concentrates on a certain point for eg if we have to find a probability of getting a number 2. Then our whole concentration is on 2. Hence we use pmf however in pdf our concentration our on the interval it is lying. For eg negative infinity <= X <= ...


0

This is not true $X \sim \text{Poi}(\mu)$ $Y \sim \text{Poi}(\lambda)$ $$P(X = Y - k) = \sum_{j = k}^\infty \frac{e^{-\lambda} \lambda^j}{j!} \frac{e^{-\mu} \mu^{j-k}}{(j-k)!} = f(\mu)$$ $$ f'(\mu)= \frac{e^{-\lambda} \lambda^k}{k!} (-e^{-\mu}) + \sum_{j = k+1}^\infty \frac{e^{-\lambda} \lambda^j}{j!}\bigg( \frac{-e^{-\mu} \mu^{j-k}}{(j-k)!}+ ...


0

(a) By symmetry $E(X_i)=\frac{1}{2}$. By independence $E(X_1X_2)=E(X_1)E(X_2)$. (b) The variance of $X_1X_2$ is $E((X_1X_2)^2)-(E(X_1X_2))^2$. It remains to calculate $E((X_1X_2)^2)$. By independence, this is $E(X_1^2)E(X_2^2)$. To calculate $E(X_i^2)$, note this is $\int_0^1 x_i^2 \,dx_i$.


2

For (b), we want $E(X^2-4XY+4Y^2)$, which is $E(X^2)-4E(XY)+4E(Y^2)$. However, we cannot say in general that $E(XY)=E(X)E(Y)$. To calculate $E(XY)$, recall perhaps that $\text{Cov}(X,Y)=E(XY)-E(X)E(Y)$. We know the covariance, and $E(X)$ and $E(Y)$, so we know $E(XY)$. For (c), as you know the $\pi$ makes no difference. We use the formula ...


1

Given that $N_m=k$, where $k\ge 1$, and $0\lt x\lt 1$, we have $$\Pr(M_m\le x)=x^k.$$ Now we need to decide what $\Pr(M_m\le x)$ is if $N_m=0$. There is no obvious definition. With not much conviction, we call this probability $1$. Then $$\Pr(M_m\le x)=\sum_{k=0}^\infty e^{-m} \frac{m^k}{k!}x^k.$$ We recognize the sum as $e^{-m}e^{mx}$.


1

A reasonable interpretation of "sum" or "multiply" for two probability distributions is the distribution of the sum or product of two independent random variables with those distributions. However, there's no natural way to have an "inverse".


1

This is not trivial. For example density function of sum of two continuous variables is convolution of densities of these variables.


1

Hint: $$ Var(X)=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2. $$ Therefore, $E[X^2]=Var(X)+E[X]^2$. Then, $$ E[X-2(X-1)^2]=E[X-2(X^2-2X+1)]=E[-2X^2+5X-2]=-2E[X^2]+5E[X]-2. $$ Now, substitute for variance and expectation.



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