Question bank — Probability mass and density functions

True — $p_X(x)=P(X=x)$ is a genuine probability, and every probability lives in $[0,1]$ by the axioms.

False — density is probability per unit length, so a narrow tall spike (e.g. Uniform on $[0,0.5]$ has $f=2$) legally exceeds 1; only the total area is capped at 1.

False — zero probability is not impossibility; a single point has zero width so zero area, yet $X$ must land on some exact value each time.

True — the two endpoints each contribute zero area, so including or excluding them changes nothing.

False — the endpoints carry positive mass when they are attainable values, e.g. for a die $P(1\le X\le 6)\ne P(1<X<6)$.

False in meaning — $p_X(x)$ is a probability you can read off directly; $f_X(x)$ is a rate you must integrate. Conflating them is the root of most PDF mistakes.

False — you also need every $p_X(x)\ge 0$; a list summing to 1 with a negative entry is not a probability model.

Mostly true for the usual cases but not required in general — integrability forces the tails' area to vanish, yet contrived densities can have non-vanishing thin spikes; for the standard distributions (Gaussian, exponential) it does tend to 0.

False — $F_X(x)=P(X\le x)$ accumulates non-negative density, so it is non-decreasing everywhere; a low density only means a gentle slope, never a drop.

False — it holds where $F_X$ is differentiable, i.e. for continuous $X$; a discrete or mixed variable has jumps in $F_X$ where no ordinary derivative exists.

Error: a die only outputs integers, so $p_X(3.5)=0$; being inside the numeric range is irrelevant — only attainable values carry mass.

Error: $f_X(0.5)$ is a density value, not a probability; the probability at a single continuous point is $0$, and you must integrate over an interval to get a number in $[0,1]$.

Error: normalisation constrains the area $\int f_X=1$, not the peak height; the peak can be anything as long as the total area works out.

Error: the compact form is only defined for $x\in \{0,1\}$; outside that set $p_X(x)=0$, and plugging in $x=2$ is meaningless.

Error: it's the opposite — a wider curve must be shorter, since the fixed total area of 1 spread over more width forces the peak down; the normaliser $\frac{1}{\sigma \sqrt{2\pi }}$ shrinks the height.

Error: that subtracts densities, which is nonsense; the interval probability is ${\int }_a^b{f}_{X}dx=F_X(b)-F_X(a)$ — you subtract CDF values, not densities.

Error: higher $\lambda$ means a steeper decay and shorter waits; the extra height near 0 reflects that arrivals happen sooner, not later.

Because discrete probability sits as concentrated lumps ("mass") at specific points, while continuous probability is smeared out and you measure its concentration per unit length ("density"), exactly like physical mass vs. mass-density.

A single continuous point has zero width and zero probability, so the density there tells you nothing usable alone; probability only emerges by summing density over a width, which the integral does.

$F_X(x)=P(X\le x)$ is defined for any $X$; it sums a PMF up to $x$ and integrates a PDF up to $x$, so the CDF is the common language where the discrete/continuous split dissolves.

$F_X$ accumulates probability, so its slope at $x$ is exactly how fast probability piles up there — that rate is the density $f_X(x)$, by the Fundamental Theorem of Calculus.

The likelihood of observing a labelled outcome under a predicted probability $p$ is exactly $p^x(1-p{)}^{1-x}$; maximising the product of these over data recovers the fitting objective for binary classifiers.

Softmax outputs one non-negative number per discrete class and they sum to 1 — that is precisely a PMF over a finite set of categories, since the outcome space is discrete labels, not a continuum.

If two densities differ only on a set of zero total width (measure zero) — say at isolated points — their integrals over every interval match, so they describe the same distribution despite differing values at those points.

It can be taken as $1$ or $0$ there; because a single point contributes zero area, the boundary value is a matter of convention and never affects any probability.

It equals $\lambda$ (its maximum), and the density is only defined for $x\ge 0$; for $x<0$ it is $0$, marking the hard start of a waiting time that cannot be negative.

It is a degenerate discrete variable with PMF $p_X(5)=1$ and $0$ elsewhere; there is no meaningful PDF because all probability is a single point mass, giving a CDF that jumps from 0 to 1 at $x=5$.

Yes — a mixed distribution (e.g. rainfall: a lump of mass at exactly $0$ mm plus a continuous density for positive amounts) has a CDF with both a jump and a smooth rise, and needs neither a pure PMF nor a pure PDF alone.

The tails must decay fast enough that their infinite extent still encloses finite area; the Gaussian's $e^{-(x-\mu {)}^2/2{\sigma }^2}$ decays faster than any polynomial, guaranteeing convergence to a finite total that the normaliser scales to exactly 1.

It is $0$ — the exponential is supported only on $x\ge 0$, so no probability mass lies below $0$; the CDF $F_X(x)$ is flat at $0$ for all negative $x$.