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For questions about probability. independence, total probability and conditional probability. For questions about the theoretical footing of probability use [tag:probability-theory]. For questions about specific probability distributions, use [tag:probability-distributions].

1 vote

Conditional probability proof of $P(B|A)$

It is simply a bracket, not the box function. The proof just requires using the inequality $P(A\cap B)\ge P(A)+P(B)-1$, which follows from the fact that $P(A\cup B)\le1$.
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Finding joint pdf of $(U,V)$, where $U$ and $V$ are transformations of independent $N(0,1)$ ...

It is clear from the definition of $U$ and $V$ that a polar transformation would be useful. Changing variables $(X,Y)\to (R,\Theta)$, you get the joint density of $(R,\Theta)$: $$f_{R,\Theta}(r,\the …
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2 votes
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Find $ \operatorname{Cov}(X^2,Y^2)$

Assuming you mean $(X,Y)$ is jointly normal where $X$ and $Y$ have zero means and unit variances and $\operatorname{Corr}(X,Y)=\rho$, we know the conditional distribution of $Y\mid X$, namely $$Y\mid …
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2 votes
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Proof of Binomial distribution asymptotic to Normal distribution.

You can directly prove that the MGF $M_{Z_n}(t)$ of the standardised binomial variate $Z_n=\frac{X_n-np}{\sqrt{npq}}$ tends to the MGF of the $\mathcal{N}(0,1)$ distribution for large $n$. I briefly s …
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Probability that a family has exactly r boys

left(\frac{p}{2}\right)^r\sum_{t=0}^\infty \binom{t+r}{t}\left(\frac{p}{2}\right)^t\quad, t=k-r$$ $$=\alpha\left(\frac{p}{2}\right)^r\left(1-\frac{p}{2}\right)^{-(r+1)}$$ For the 2nd part the required probability
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There are $2^n$ tickets in a jar.The frequency of the tickets of number $i$ is ${n \choose i}$

Without resorting to generating functions we can do the following: Let $X_i$ denote the number on the $i$th ticket for $i=1,2,...,m$, so that $S=\sum\limits_{i=1}^mX_i$. The pmf of $X_i$ for all $i= …
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Finding PMF from CDF for 2 Variables

If $F(a,b)=P(X\le a,Y\le b)$ is the DF of the random vector $(X,Y)$, then the amount of jump of $F$ at $(a,b)$ is given by $$P(X=a,Y=b)=F(a,b)-F(a-0,b)-F(a,b-0)+F(a-0,b-0)$$ where $F(a-0)=\lim_{h\to0 …
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Find the conditional expectation of two correlated standard normal

Just to get you started. By definition, $$E[X\mid Y>0]=\frac{E[X\mathbf1_{Y>0}]}{P(Y>0)}=\frac{1}{P(Y>0)}\iint x\mathbf1_{y>0} f_{X,Y}(x,y)\,dx\,dy$$ You can see this from the conditional density …
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3 votes

Let $X_1$ and $X_2$ be i.i.d. normal can we find the distribution of $(U_1,U_2)=(\max(X_1,X_...

Suppose $X_{(1)}=\min(X_1,X_2)$ and $X_{(2)}=\max(X_1,X_2)$. Let $f$ and $F$ denote the density and distribution function of the (normal) population. Then for $x<y$, the distribution function of $(X_ …
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How to find ratio of two Continuous random variables

\frac{\max(X,1-X)}{\min(X,1-X)}=\begin{cases}\frac{X}{1-X}&,\text{ if }X\geqslant 1/2\\\\\frac{1-X}{X}&,\text{ if }X<1/2\end{cases} \end{align} So, purely mechanically, we can write (using the total probability
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Consider three events, $A$ ,$B$ and $C$. $P(A) = 0.7$ and that $P(B) = 0.63, P(C) = 0.35$

Hint: You can write $$A\cup B=A\cup (B-A)$$, so that when you apply the probability function on $A\cup B$ you have the union of two disjoint events, namely $A$ and $B-A$. …
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Let X and Y be two independent Bernoulli variables of parameter $\frac{1}{3}$

$(X,Y)$ are independent $\text{Ber}(p)$ variables with $p=1/3=1-q$. You have \begin{align}\Pr(V=0,U=0)&=\Pr(X+Y=0,XY=0)\\&=\Pr(X=0,Y=0)\\&=\Pr(X=0)\Pr(Y=0)\\&=q^2\end{align} and \begin{align}\Pr(U=0 …
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Confusion when finding joint density of $Y-X$

You have incorrectly taken $x$ from $+z$ to $\infty$ in the outer integral; rather we have $\color{red}-z<x<\infty$. Because $$y>0,y-x\le z<0\implies x\ge y-z> -z$$ For each $z<0$, the distribution …
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2 votes
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Computing conditional expectation for a Poisson process

Using the fact that $N_t-N_s$ is independent of $N_s$ for $0\le s<t$ (a defining property of the Poisson process $(N_t)_{t\ge 0}$), one gets $$E(N_t-N_s)=E(N_t-N_s\mid N_s)=E(N_t\mid N_s)-N_s$$ Theref …
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Let $X,Y,Z$ be independent $N(0,1)$ variables, $R = X^2 + Y^2 + Z^2$, find the probability d...

Sorry to say, but your work doesn't make much sense. You should be working with densities, not probabilities. More explicitly, using change of variables to find the distributions as it looks like the …
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