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Yes so when you roll one dice and you would like to obtain a $2$, the chance is $\frac{1}{6}$, when there are two die, there are now $36$ possible outcomes which you can list for example $(1,1) (1,2)..... (6,6)$. For this there are $11$ outcomes, try and write out yourself it will make more sense! So the chance is $\frac{11}{36}$ which is greater than only ...


2

We have $$ P(X = k | X+Y = m) = \frac{P(X = k \,\&\, Y = m-k)}{P(X+Y = m)} = \frac{p_{k}p_{m-k}}{\sum_{j=0}^{m}p_{j}p_{m-j}} = \frac{1}{m+1} $$ for all $0\leq k \leq m+1$. In particular, we have $$ p_{0}p_{m} = p_{1}p_{m-1} = p_{2}p_{m-2} = \cdots = p_{m}p_{0}. $$ This gives $$ \frac{p_{1}}{p_{0}} = \frac{p_{m}}{p_{m-1}} $$ for all $m\geq 1$, so that $...


2

$P(Y<1|X<1)=\frac {P(X<1,Y<1)} {P(X<1)}=\frac {\int_0^{1} \int_x^{1}2e^{-x-y} dydx} {\int_0^{1} \int_x^{\infty}2e^{-x-y} dydx}$. I will let you evaluate the integrals.


2

So the log likelihood given $n$ observations is $$l(\theta) = \log \prod_{i=1}^n f(y_i;\theta) \propto - \sum_{i=1}^n \log y_i - \frac{n}{2} \log \theta - \frac{1}{2\theta}\sum_{i=1}^n \log^2 y_i$$ Derive w.r.t $\theta$ and set it to zero to get the first order condition: $$l'(\theta) = - \frac{n}{2\hat{\theta}} + \frac{1}{2\hat{\theta}^2} \sum_{i=1}^n \log^...


2

If $m < n$ and $f$ is everywhere differentiable, then measure of $f(\mathbb R ^ m)$ is zero (see, for example, this answer - we can extend $f$ to $\mathbb R^n$ to apply it directly by using $g(x_1, \ldots, x_n) = f(x_1, \ldots, x_m)$ - then $\det D g$ is zero everywhere). So support of $f(X)$ has zero measure and thus $f(X)$ doesn't have density.


2

Your intuition for $100$ dice is correct – the probability that at least one two appears in $n$ die rolls is $1-\left(\frac56\right)^n$, where $\frac56$ is the probability of not getting a two from one of the dice. $1-\left(\frac56\right)^n$ tends to one as $n$ increases. For two dice the probability of at least one two works out as $\frac{11}{36}>\...


2

Let's assume $n$ die are thrown and we want the probability of at least $1$ dice to show $2$. $$P(\text{at least one 2 shows up}) = 1- P(\text{No 2 shows up})$$ $$=1-\left(\frac{5}{6}\right)^n$$ For $n=100$, the probability comes to be around $0.999999$. So yes it's very likely when you throw a $100$ die, atleast one $2$ will show up.


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The PDF tells us that $X<Y$ a.s. so that $P(Y<1\mid X=1)=0$.


2

As we have $0<x<y<\infty$, if we take $x$ as dependent on $y$, we get - $$\iint_R f_{XY}dx dy= \int_0^\infty\int_0^yf_{XY}dxdy$$ If we would take $y$ as dependent on $x$, we get - $$\iint_R f_{XY}dx dy= \int_0^\infty\int_x^\infty f_{XY}dydx$$


2

Yes, this is fine. To add some detail: for the second equality, you've used that $\int_y p(x,y)\,dy = 1$ for all $x$. The third is justified since $f(x)$ is just a constant as far as $\int_y$ cares. And for the last line, to move to the joint integral you might invoke Fubini-Tonelli.


2

We seek to find the number of solutions to $x_1 + x_2 + \ldots + x_k =n$ in nonnegative integers. Consider $n$ dots and $k-1$ dividers arranged in a row. The dividers divide the row into $k$ sections, and the number of dots in each gives the $x_i$. There are $n+k-1 \choose k-1$ ways to arrange these, and so this is the number of ways votes can be cast.


2

You have $n$ votes for $k$ candidates. If you line up the votes by candidate, you'll have groups of votes for each one. (Unpopular candidates may not have any votes.) We can insert $k-1$ dividers in the line of votes to cordon them off by candidate. So if we have $n+k-1$ spaces, we'll have room for all of the votes and the dividers. Now the problem gets ...


1

Suppose $X_i$ are iid with continuous cdf $F$, $L_n = \min(X_1,\ldots,X_n)$ and $U_n = \max(X_1,\ldots,X_n)$. Then $$\mathbb P(a \le L_n \le U_n \le b) = \mathbb P( X_1,\ldots,X_n \in [a,b]) = (F(b) - F(a))^n$$ and if $F$ corresponds to a pdf $f$, the joint pdf of $(L_n, U_n)$ is $$f_{L_n, U_n}(x,y) = - \dfrac{\partial^2}{\partial x \partial y} (F(y) - F(x)...


1

$EX^{-2}=100\int_0^{\infty} \frac 1 x e^{-10x}\, dx$. Since $e^{-10x} \to 1$ as $x \to 0$ and $\int_0^{1} \frac 1 x \, dx=\infty$ it follows that $EX^{-2}=\infty$.


1

$f_X(x)=\int_x^{\infty} 2e^{-x-y} dy=2e^{-x}e^{-x}=2e^{-2x}$ for $0<x<\infty$. Also $F_X(x)=\int_0^{x} 2e^{-2t} dt =-e^{-2t}|_0^{x}=1-e^{-2x}, 0<x<\infty$.


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For any $\omega$ either $ \omega \in A$ or $ \omega \in A^{c}$. In the first case both sides of the equation are $g(x)$ and in the second case both sides are $g(y)$ so the equation is true. $g(\alpha X+\beta Y)=g(\alpha)X+g(\beta)Y$ is not true in general.


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Hint: $X$ also has normal distribution. Since $\beta^{t}=e^{ta}$ where $a =\log\, \beta$ $X$ has same distribution as $Y+a$ where $Y$ has MGF $e^{t^{2}}$. Now you can use normal density function to evaluate the probability.


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Using, as in the linked post, $p=\frac \lambda n$ $$A=p^k \binom{n}{k}p^k (1-p)^{n-k}=\binom{n}{k} \left(\frac{\lambda }{n}\right)^k \left(1-\frac{\lambda}{n}\right)^{n-k}$$ Taking logarithms and expanding as a Taylor series for large values of $n$ to get $$\log(A)=\left(k \log (\lambda )+\log \left(\frac{e^{-\lambda }}{k!}\right)\right)+\frac{-k^2-\lambda ^...


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The conditional density you have found is for $0<x<y$. $E(Y|X)=\int_x^{\infty} ye^{x-y} dy$. Integrate by parts to find the exact value.


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There are $11 \choose 7$ options for which $7$ attempts have been successful, each with $\left( \frac{1}{4} \right) ^ 7 \left( \frac{3}{4} \right)^4$ change of occurring. So the probability of $A$ occurring $7$ out of $11$ times is ${11 \choose 7} \left( \frac{1}{4} \right) ^ 7 \left( \frac{3}{4} \right)^4=0.637\%$. In general, you use the binomial ...


1

If you can convert the percentages of the population to integers (as in your example): Say there are $c=N\cdot\frac{x}{100}$ individuals in the population with the characteristic; and the sample size is $k=N\cdot \frac{y}{100}$. Then the probability of at least one person having the characteristic would be $$1-\frac{\binom{N-c}{k}}{\binom{N}{k}}$$ The ...


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This is conditional expected probability. Given that you scratched a bunch of tickets, you know that the probability of the last one being a winner is increased only because you know that the tickets are not independently random; there is a set $1000$ out of a million winners. However, if each ticket was independently random (i.e each ticket had a $0.1\%$ ...


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