Question bank — Joint distributions — joint PMF - PDF, marginal, conditional

False — it is a density (probability per unit area). It can even exceed $1$; only ${\iint }_{A}fdxdy$ over a region is a probability.

True — factoring of the joint into the product of marginals for every point is exactly the definition of Independence of Random Variables.

False — marginals describe each variable alone and say nothing about their joint behaviour; equal marginals can coexist with strong dependence.

False — many different joints share the same two marginals (they differ in how the variables move together), so the pair of marginals does not pin the joint down.

True — the conditional is renormalized by dividing by $f_Y(y)$, so as a function of $x$ it integrates to exactly $1$; it is a genuine distribution in $x$.

False — integrating the conditional over $y$ as well double-counts; it need not equal $1$. Only the integral over $x$ (with $y$ frozen) equals $1$.

False — zero Covariance and Correlation means no linear association; nonlinear dependence (e.g. $Y=X^2$ with symmetric $X$) can give zero covariance yet full dependence.

True — independence makes $E[XY]=E[X]E[Y]$, so covariance vanishes. (The converse fails, as above.)

True — dividing the joint entry by $p_Y(y)\le 1$ inflates it, so the conditional probability is at least as large as the joint value.

True in form, but for a fixed $y$ it is a distribution in $x$; different conditioning values $y$ generally give different-shaped slices.

Wrong — deleting keeps one $Y$-value; marginalizing means summing over all $Y$-values: $p_X(x)={\sum }_y{p}_{X,Y}(x,y)$. You collapse, never drop.

Wrong — that slice integrates to $f_Y(y)$, not $1$. You must divide by $f_Y(y)$ to renormalize it into a valid distribution.

Wrong bounds — on this triangle $y$ runs only from $0$ to $x$, so $f_X(x)={\int }_0^{x}3xdy=3x^2$. Ignoring the support gives the wrong marginal.

Wrong — the support $0<y<x<1$ couples the variables (the allowed range of $y$ depends on $x$). Factoring must hold including a product-shaped support; here it does not.

Wrong — both use $p_{X,Y}(x,y)$ on top but divide by different marginals ($p_Y(y)$ vs $p_X(x)$), so they are generally unequal (this is the seed of Bayes' Theorem).

Wrong — a tall narrow density (e.g. a spike on a short interval) exceeds $1$ in height while still enclosing unit area. Height is unbounded; only total area is fixed.

Wrong — dividing by $0$ is undefined; the conditional $f_{X\mid Y}(x\mid y)$ is only defined where $f_Y(y)>0$, i.e. where that value of $Y$ actually carries density.

Because a single point $(x,y)$ has zero area, so $P(X=x,Y=y)=0$; density measures probability per unit area, recovering real probability only when integrated over a region.

Because $\{X=x\}={\bigcup }_y\{X=x,Y=y\}$ is a union of disjoint events; additivity of probability turns that union into a sum (discrete) or integral (continuous) — this is the Law of Total Probability.

The frozen slice $\{Y=y\}$ has total mass $p_Y(y)$ (or $f_Y(y)$), not $1$; dividing rescales it so the new sample space integrates to $1$ and is a legitimate distribution.

Under independence, knowing $Y$ tells you nothing about $X$, so $f_{X\mid Y}=f_X$; substituting into the chain rule $f_{X,Y}=f_{X\mid Y}{f}_Y$ gives the product form.

Both expressions reconstruct the same joint from a slice times its weight; conditioning on $X$ first or $Y$ first are two routes to the identical surface — equating them yields Bayes' rule.

Independence needs the joint to factor as $g(x)h(y)$ over a rectangular support; a triangular or curved support makes the allowed values of one variable depend on the other, which is dependence.

Covariance depends on $E[XY]$, a joint quantity; the marginals fix $E[X]$ and $E[Y]$ but many joints with those marginals give different $E[XY]$.

It is left undefined — there is no probability mass to condition on, so the $0/0$ ratio has no meaning there.

Only $x\in (y,1)$ — freezing $Y=y$ forces $x>y$, so the conditional lives on the shrunken interval, not all of $(0,1)$.

Yes — a constant carries no randomness, so $p_{X,Y}(x_0,y)=1\cdot p_Y(y)=p_X(x_0)p_Y(y)$; the factorization holds trivially.

It means $p_X(x)=0$ (that $X$-value never occurs); the conditional $p_{X\mid Y}(x\mid y)=0$ for every $y$, and conditioning on $X=x$ is undefined.

That $X$'s distribution does not react to $Y$, i.e. $f_{X\mid Y}=f_X$ — the definition of independence.

Exactly $0$ — the "and $0$ elsewhere" clause is part of the density; forgetting it wrongly extends integration bounds and breaks the total-mass-$=1$ check.