22 votes

A and B are independent. Does P(A ∩ B|C) = P(A|C) · P(B|C) hold?

No this is not in general true, as you can see from a simple counter example: Toss two independent coins. Event $A$ is coin 1 head. $P(A)=0.5$ Event $B$ is coin 2 head. $P(B)=0.5$ Event $C$ is either ...
George Savva's user avatar
  • 2,034
20 votes
Accepted

Are linear combinations of independent random variables again independent?

Yes, for the content of your question. and No, for the title, in general. Yes: Your $a_1,\dots, a_n$ are just some constant numbers. Then the independence of $X_1,\dots, X_n$ implies that the $...
Ute's user avatar
  • 2,510
15 votes
Accepted

Probability that X > Y when X ~ N(0,2) and Y ~ N(0,1)

The fact that the distribution of $X-Y$ has mean $0$ does NOT imply that $\mathbb{P}(X-Y<0)$ and $\mathbb{P}(X-Y>0)$ are equal. So for the last step it is not enough that $X-Y$ has mean $0$, you ...
user133281's user avatar
8 votes

Are linear combinations of independent random variables again independent?

If $X_1,...,X_n$ are mutually independent then $a_1 X_1,...,a_n X_n$ are also mutually independent (i.e., using scalar multiples does not get rid of independence). However, the quantity $Y_n = \sum ...
Ben's user avatar
  • 123k
8 votes

Probability that X > Y when X ~ N(0,2) and Y ~ N(0,1)

Your reasoning seems right. You can always do some simulation to check if your answer is correct. ...
Peter Flom's user avatar
  • 117k
7 votes
Accepted

Are SHAP values potentially misleading when predictors are highly correlated?

Summary Yes, SHAP values are potentially misleading when predictors are correlated -- they can be imprecise and even have the opposite sign. The correlation does not need to be incredibly high, around ...
Dudelstein's user avatar
7 votes

Different amounts of observations for individuals - Which model to use?

As I mentioned in the comment, mixed/hierarchical/multilevel models are the general way to address this form of non-independence. The model is modified to deal with the correlations between points ...
mkt's user avatar
  • 18.1k
6 votes
Accepted

Show that the two random variables with F-distribution are independent

Your answer to (a) is correct. For part (b), the two random variables are $F$-distributed by construction. We could prove that they're independent by establishing joint independence of $Y_1, Y_2, X_3$....
Doctor Milt's user avatar
  • 2,672
6 votes

Prove that two random variables are independent

Conditioned on $X$ having value $x$, the distribution of $Y$ is $N(x,x^2)$. The conditional distribution of $Z = \dfrac YX$ given that $X=x$ is the same as the distribution of $\dfrac Yx$ which, as ...
Dilip Sarwate's user avatar
6 votes

Different amounts of observations for individuals - Which model to use?

If your sample data is representative of what your actual data will look like, then I don't think multilevel models are the way to go. First, most of your people have one observation and only one has ...
Peter Flom's user avatar
  • 117k
6 votes

Statistical test for nominal data, multiple variables and within-subject design in R

You have a binary response (osteophytes), for which logistic regression is a good general approach. You have non-independence due to multiple measurements on the ...
mkt's user avatar
  • 18.1k
5 votes
Accepted

What is the conditional $\operatorname{Var}(XY|Y)$ given that $X$ and $Y$ are independent?

Your reasoning flow shows that you had a good understanding on conditioning. Formally, you can derive it from the definition$^\dagger$ of conditional variance and basic properties of conditional ...
Zhanxiong's user avatar
  • 17.2k
5 votes

Representation of two Gaussian vectors as sums of independent Gaussian vectors

Let $X$ be $n$-dimensional, $Y$ be $m$-dimensional, and both with zero means. (It's simple to deal with nonzero means later because they just get added in.) We seek an $n$-dimensional zero-mean ...
whuber's user avatar
  • 321k
5 votes
Accepted

Conditional distribution $f(x|y)$ if $X$ and $Y$ are independent

Yes, if the conditional density does not depend on the input $y$ then that is sufficient for independence. To see this, suppose that we can write $f(x|y) = h(x)$ for some function $h$ (i.e., as a ...
Ben's user avatar
  • 123k
4 votes
Accepted

Let $X_1,X_2,\ldots$ be iid random variables with Cauchy distribution and $S_n=X_1+X_2+\cdots+X_n$, find $P(S_n>an)$, $a>0$

If $X_1, \ldots, X_n$ are i.i.d. standard Cauchy distribution $C(0, 1)$, then since Cauchy distribution is closed under independent summation (see the second item of this link), the distribution of $...
Zhanxiong's user avatar
  • 17.2k
4 votes
Accepted

If a strictly stationary process is also independent, does this imply i.i.d.?

Yes, that is correct. Strict stationarity implies a common marginal distribution for the variables in the series, which is the ID part in IID. If you combine this with an assumption of independence ...
Ben's user avatar
  • 123k
4 votes

Prove that two random variables are independent

You have $$f_{U,V}(u,v) =\frac{v}{\sqrt{2\pi}v^2}e^{-\frac{1}{2v^2}(uv-v)^2}I_{\mathscr Y}((u,v))$$ though I think you should extend the square root to $\dfrac{v}{\sqrt{2\pi v^2}}e^{-\frac{1}{2v^2}(...
Henry's user avatar
  • 38.5k
4 votes

Probability that X > Y when X ~ N(0,2) and Y ~ N(0,1)

I'd like to add another simulation implemented in R. First, we can define PDFs for either random variable: ...
user654123's user avatar
3 votes
Accepted

Does strict monotonicity imply image variable is dependent with domain variable?

No -- there are simple counterexamples. I take it that $X:(\Omega,\mathfrak F, \mathbb P)\to (E,\mathcal E)$ is a (generalized) random variable, that $E$ is endowed with a partial order $\le,$ and ...
whuber's user avatar
  • 321k
3 votes
Accepted

Chi squared and Cramer's V as a measure of independence

I think your confusion comes from several things. In the first place, your calculations are theoretically correct, but apply to a two-way contingency table. The rub here seems to be that it looks like ...
J-J-J's user avatar
  • 3,852
3 votes

Are SHAP values potentially misleading when predictors are highly correlated?

By default, TreeExplainer in the SHAP (SHapley Additive exPlanations) library uses feature_perturbation = "interventional". This choice is based on the ...
Mika Rafieferantsoa 's user avatar
3 votes

Show that the two random variables with F-distribution are independent

To show the independence of $\frac{X_1/r_1}{X_2/r_2}$ and $\frac{X_3/r_3}{(X_1 + X_2)/(r_1 + r_2)}$ in part (b), denote $\frac{X_1/r_1}{X_2/r_2}$ by $Z_1$, $\frac{X_1 + X_2}{r_1 + r_2}$ by $Z_2$, and $...
Zhanxiong's user avatar
  • 17.2k
3 votes

Quick question about density with respect to product measure

For a counterexample let $\Omega_1 = \Omega_2 = [0,1]$ $\mathcal A_1 = \mathcal A_2 = \mathcal B_1 = \mathcal B_2 = \mathcal B([0,1]) \equiv \mathcal B(\mathbb R)\big|_{[0,1]}$ $\mathbb P_1 = \mathbb ...
statmerkur's user avatar
  • 5,810
3 votes

Case when random variable $X$ and its square $X^2$ are independent

I want to add that a sufficient and necessary condition for $X$ and $X^2$ are independent is that $X^2$ is degenerate. The sufficiency is trivial. Conversely, suppose $X$ and $X^2$ are independent. ...
Zhanxiong's user avatar
  • 17.2k
3 votes
Accepted

Calculate joint distribution from marginal distributions

Yours is a particular case of the following theorem. Theorem. Let $X\,\sim\, \text{N}_p(\mu, \Sigma)$, $\underset{q\times p}{A}$ a and $\underset{q\times 1}{c}$ a fixed matrix and a fixed vector, ...
utobi's user avatar
  • 11.6k
3 votes
Accepted

Can you combine effects from both the composite and domain levels of questionnaires in a meta-analysis?

I would say, yes, one can do this. However, I would not include subdomain level results and the composite result from the same study, since most 'composites' are just sum scores or means based on the ...
Wolfgang's user avatar
  • 16.9k
3 votes

Independence across observations

Consider this counterexample showing that independence can hold without random sampling: Let $N_1, N_2 \sim \mathcal{N}(0, 1)$ independently, and let $X_1 := N_1$ and $X_2 := \beta X_2 + N_2$. Given ...
Scriddie's user avatar
  • 2,084
3 votes
Accepted

Independence across observations

Short answer: No. Longer answer: Random sampling is a different thing from independence. Take the archeypical sample of "sophomore college students in a survey course". This is not a random ...
Peter Flom's user avatar
  • 117k
2 votes

Can we conclude from $\DeclareMathOperator{\E}{\mathbb{E}}\E g(X)h(Y)=\E g(X) \E h(Y)$ that $X,Y$ are independent?

This is almost Proposition 7.1.3(i) from Athreya & Lahiri 2006 sans some minor differences in formatting: Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $\{X_1, \ldots, X_k \}$, $2 ...
Galen's user avatar
  • 8,095
2 votes

T-test with multiple measures per subject

No, probably not OK. How large an effect would it be reasonable to suppose that a green card would have on reported happiness? I would be surprised if it caused even a minor change and that means that ...
Michael Lew's user avatar
  • 14.8k

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