Quiz 3

Week 9: Random Variables, PMF & CDF

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2026-09-12T00:00:00.000Z
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Random variables mapping S to R, discrete vs continuous, Probability Mass Function (PMF), and Cumulative Distribution Function (CDF). # Statistics for Data Science I — Week 9: Random Variables, PMF & CDF IITM’s Week 9 introduces **random variables, random experiments and sample spaces, discrete and continuous random...

Statistics for Data Science I — Week 9: Random Variables, PMF & CDF

IITM’s Week 9 introduces random variables, random experiments and sample spaces, discrete and continuous random variables, probability mass functions, and cumulative distribution functions.

TL;DR

Until now, probability dealt directly with events.
Week 9 introduces a function:
X:S→R\boxed{ X:S\to\mathbb R }
called a random variable.
It assigns a numerical value to each outcome.
Example:
Two coin tosses:
S={HH,HT,TH,TT}S=\{HH,HT,TH,TT\}
Define:
X=number of heads.X=\text{number of heads}.
Then:
X(HH)=2X(HH)=2 X(HT)=1X(HT)=1 X(TH)=1X(TH)=1 X(TT)=0.X(TT)=0.
From there we study the distribution of XX rather than the raw sample-space outcomes.

1. Random experiment recap

A random experiment is a process with uncertain outcome.
Examples:
  • toss coins
  • roll dice
  • count arrivals
  • measure waiting time
  • observe rainfall
Every experiment has a sample space:
S.S.

2. Random variable

A random variable is a function assigning a real number to each outcome:
X:S→R\boxed{ X:S\to\mathbb R }
Despite the name, the function itself is not "random."
The randomness comes from which outcome in SS occurs.

3. Example: one coin

Sample space:
S={H,T}.S=\{H,T\}.
Define:
X(H)=1X(H)=1 X(T)=0.X(T)=0.
Then XX is a numerical encoding of the random experiment.

4. Example: two coins

Let:
X=number of heads.X=\text{number of heads}.
Then:
OutcomeXX
HH2
HT1
TH1
TT0
Possible values of XX:
{0,1,2}\boxed{ \{0,1,2\} }
Notice:
44
sample-space outcomes collapse into:
33
possible random-variable values.

5. Why random variables are useful

Raw outcomes can be complicated.
Example:
HHHTHTH…HHHTHTH\ldots
But often we only care about a numerical feature:
  • number of heads
  • total sales
  • waiting time
  • number of failures
  • temperature
  • profit
So random variables compress the experiment into a quantity we can analyze.

6. Capital vs lowercase notation

Typically:
XX
denotes the random variable.
A specific observed value is written:
x.x.
For example:
X=3X=3
means the random variable takes value 33.

7. Event generated by a random variable

Statements such as:
X=2X=2
represent events.
Example:
If XX is number of heads in two tosses:
{X=2}={HH}.\{X=2\} = \{HH\}.
Similarly:
{X≥1}={HH,HT,TH}.\{X\geq1\} = \{HH,HT,TH\}.

8. Probability notation

We write:
P(X=x)P(X=x)
to mean:
probability that random variable XX takes value xx.
Example:
P(X=2)=P(HH)=14.P(X=2)=P(HH)=\frac14.

9. Discrete random variable

A random variable is discrete if it takes values in a finite or countable set.
Examples:
  • number of heads
  • number of defects
  • number of customers
  • die result
  • number of goals
Possible values might be:
0,1,2,3,…0,1,2,3,\dots

10. Continuous random variable

A random variable is continuous when its possible values form a continuum.
Examples:
  • height
  • waiting time
  • temperature
  • distance
  • lifetime of a component
Possible values can lie throughout an interval:
[a,b][a,b]
or:
[0,∞).[0,\infty).

11. Discrete vs continuous

Discrete

Usually associated with counting.
Example:
X=number of calls.X=\text{number of calls}.

Continuous

Usually associated with measurement.
Example:
X=call duration.X=\text{call duration}.
This mirrors the data-type distinction from Week 1.

12. Probability distribution

A probability distribution describes:
which values the random variable can take, and how probable those values are.
For a discrete variable, we often display:
xxP(X=x)P(X=x)
00.25
10.50
20.25
The probabilities must satisfy:
P(X=x)≥0P(X=x)\geq0
and:
∑xP(X=x)=1\boxed{ \sum_xP(X=x)=1 }

13. Probability mass function

For a discrete random variable, the probability mass function or PMF is:
pX(x)=P(X=x)\boxed{ p_X(x)=P(X=x) }
It assigns a probability mass to every possible value.

14. PMF conditions

A valid PMF must satisfy:
pX(x)≥0\boxed{ p_X(x)\geq0 }
for every xx, and:
∑xpX(x)=1\boxed{ \sum_xp_X(x)=1 }
These two checks are fundamental.

15. Example: fair die

Let:
X=number shown on a fair die.X=\text{number shown on a fair die}.
Then:
X∈{1,2,3,4,5,6}.X\in\{1,2,3,4,5,6\}.
PMF:
pX(x)=16p_X(x)=\frac16
for:
x=1,…,6.x=1,\dots,6.
Otherwise:
pX(x)=0.p_X(x)=0.

16. Piecewise PMF notation

We can write:
pX(x)={16,x∈{1,2,3,4,5,6}0,otherwisep_X(x)= \begin{cases} \frac16,&x\in\{1,2,3,4,5,6\}\\ 0,&\text{otherwise} \end{cases}
This is a common mathematical form.

17. Example: number of heads in two tosses

For fair independent tosses:
X∈{0,1,2}.X\in\{0,1,2\}.
Probabilities:
P(X=0)=14P(X=0)=\frac14 P(X=1)=24=12P(X=1)=\frac24=\frac12 P(X=2)=14.P(X=2)=\frac14.
So:
pX(x)={14,x=012,x=114,x=20,otherwisep_X(x)= \begin{cases} \frac14,&x=0\\ \frac12,&x=1\\ \frac14,&x=2\\ 0,&\text{otherwise} \end{cases}

18. Event probabilities from PMF

Suppose we want:
P(X≥1).P(X\geq1).
Then:
P(X≥1)=P(X=1)+P(X=2).P(X\geq1) = P(X=1)+P(X=2).
Thus:
=12+14=34.= \frac12+\frac14 = \boxed{\frac34}.
For discrete RVs, probabilities over a set of values are found by adding PMF values.

19. Example

Suppose:
xxpX(x)p_X(x)
00.1
10.2
20.4
30.3
Find:
P(1≤X≤2).P(1\leq X\leq2).
Then:
P(X=1)+P(X=2)P(X=1)+P(X=2) =0.2+0.4=0.2+0.4 =0.6.=\boxed{0.6}.

20. Solving for unknown PMF constants

Suppose:
pX(x)=kxp_X(x)=kx
for:
x=1,2,3.x=1,2,3.
Since total probability is 1:
k(1)+k(2)+k(3)=1.k(1)+k(2)+k(3)=1.
So:
6k=1.6k=1.
Thus:
k=16\boxed{ k=\frac16 }

21. Always normalize

The standard strategy for an unknown PMF constant is:
∑xpX(x)=1\boxed{ \sum_xp_X(x)=1 }
Then solve for the parameter.

22. Cumulative distribution function

The CDF of a random variable XX is:
FX(x)=P(X≤x)\boxed{ F_X(x)=P(X\leq x) }
This definition works for:
  • discrete RVs
  • continuous RVs
  • mixed distributions
So the CDF is more universal than the PMF.

23. CDF intuition

The CDF asks:
How much probability has accumulated up to xx?
Hence the word:
cumulative.

24. Discrete CDF example

Suppose:
xxP(X=x)P(X=x)
00.2
10.3
20.4
30.1
Then:
F(0)=P(X≤0)=0.2.F(0)=P(X\leq0)=0.2. F(1)=0.2+0.3=0.5.F(1)=0.2+0.3=0.5. F(2)=0.9.F(2)=0.9. F(3)=1.F(3)=1.

25. CDF as accumulated PMF

For discrete XX:
FX(x)=∑t≤xpX(t)\boxed{ F_X(x) = \sum_{t\leq x}p_X(t) }
So:
PMF = individual masses
CDF = accumulated mass

26. Shape of a discrete CDF

A discrete CDF looks like a staircase.
Conceptually:
text
F(x)
1.0 |                ________
    |          ______|
0.5 |    _____|
    |___|
0.0 +-----------------------> x
It jumps at values where:
P(X=x)>0.P(X=x)>0.

27. Size of a CDF jump

For a discrete variable:
P(X=x)=F(x)−F(x−)\boxed{ P(X=x) = F(x)-F(x^-) }
where:
F(x−)F(x^-)
means the CDF value immediately to the left of xx.
So a PMF value corresponds to a jump size.

28. Basic properties of a CDF

Every valid CDF satisfies:
0≤F(x)≤1\boxed{ 0\leq F(x)\leq1 }
It is nondecreasing:
x1<x2⇒F(x1)≤F(x2).x_1<x_2 \Rightarrow F(x_1)\leq F(x_2).
And:
lim⁡x→−∞F(x)=0\boxed{ \lim_{x\to-\infty}F(x)=0 } lim⁡x→∞F(x)=1\boxed{ \lim_{x\to\infty}F(x)=1 }

29. Why CDF cannot decrease

If:
x1<x2,x_1<x_2,
then:
{X≤x1}⊆{X≤x2}.\{X\leq x_1\} \subseteq \{X\leq x_2\}.
Therefore:
P(X≤x1)≤P(X≤x2).P(X\leq x_1) \leq P(X\leq x_2).
Hence FF must be nondecreasing.

30. Probability from CDF

For:
a<b,a<b, P(a<X≤b)=F(b)−F(a)\boxed{ P(a<X\leq b) = F(b)-F(a) }
This is one of the most useful CDF rules.

31. Why?

Since:
{X≤b}\{X\leq b\}
can be split into:
{X≤a}\{X\leq a\}
and:
{a<X≤b},\{a<X\leq b\},
we get:
F(b)=F(a)+P(a<X≤b).F(b) = F(a)+P(a<X\leq b).
Rearrange.

32. Discrete endpoint care

For discrete variables:
P(a≤X≤b)P(a\leq X\leq b)
may differ from:
P(a<X≤b)P(a<X\leq b)
because:
P(X=a)P(X=a)
can be positive.
So endpoint symbols matter.

33. Continuous RV endpoint intuition

For a continuous random variable:
P(X=a)=0.P(X=a)=0.
Therefore:
P(a<X<b)P(a<X<b) =P(a≤X<b)= P(a\leq X<b) =P(a<X≤b)= P(a<X\leq b) =P(a≤X≤b).= P(a\leq X\leq b).
For continuous distributions, individual points have zero probability.

34. Important continuous-variable fact

A continuous random variable can take infinitely many values, but:
P(X=x)=0\boxed{ P(X=x)=0 }
for every single exact value xx.
This initially feels strange.
It means probability is assigned to intervals, not individual points.

35. Zero probability does not mean impossible

For a continuous random variable:
P(X=1.234567)=0,P(X=1.234567)=0,
yet the observed value could theoretically equal that number.
In continuous probability:
probability zero is not identical to logical impossibility.
This distinction becomes important later.

36. PMF does not apply to continuous RVs

PMF is for discrete random variables.
For continuous variables we later use a probability density function or PDF.
The course formally introduces continuous random variables in Week 12.
For now, understand the distinction.

37. CDF works for both

Discrete:

F(x)=P(X≤x)F(x)=P(X\leq x)

Continuous:

F(x)=P(X≤x)F(x)=P(X\leq x)
Same definition.
That is why the CDF is a universal representation of a distribution.

38. Random variable transformation intuition

Suppose:
X=die resultX=\text{die result}
and define:
Y=X2.Y=X^2.
Then:
YY
is also a random variable.
Possible values:
1,4,9,16,25,36.1,4,9,16,25,36.
A function of a random variable creates another random variable.

39. Different outcomes can give same RV value

Example:
Two dice.
Let:
X=sum.X=\text{sum}.
Then:
(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
all produce:
X=7.X=7.
So:
P(X=7)=636=16.P(X=7) = \frac6{36} = \frac16.
This is a common PMF construction pattern.

40. PMF of two-dice sum

For two fair dice:
X∈{2,3,…,12}.X\in\{2,3,\dots,12\}.
Number of ways:
text
2  → 1
3  → 2
4  → 3
5  → 4
6  → 5
7  → 6
8  → 5
9  → 4
10 → 3
11 → 2
12 → 1
Therefore:
P(X=x)=number of pairs producing x36.P(X=x) = \frac{\text{number of pairs producing }x}{36}.
The PMF is not uniform even though the underlying 36 ordered die outcomes are uniform.

41. Random variable vs outcome

Do not confuse:
ω\omega
with:
X(ω).X(\omega).
Example:
Outcome:
ω=(3,4).\omega=(3,4).
If XX is sum of two dice:
X(ω)=7.X(\omega)=7.
Outcome and random-variable value are different objects.

42. Random variable vs probability

A random variable:
XX
is not itself a probability.
For example:
X=3X=3
could be a possible count.
Probability is:
P(X=3).P(X=3).
Do not confuse the numerical outcome with its likelihood.

43. PMF vs CDF

This distinction must be clear.

PMF

p(x)=P(X=x)p(x)=P(X=x)
Answers:
What is the probability at exactly this discrete value?

CDF

F(x)=P(X≤x)F(x)=P(X\leq x)
Answers:
What is the probability accumulated up to this value?

44. Example converting PMF to CDF

PMF:
xxp(x)p(x)
10.2
20.5
30.3
Then:
F(x)={0,x<10.2,1≤x<20.7,2≤x<31,x≥3F(x)= \begin{cases} 0,&x<1\\ 0.2,&1\leq x<2\\ 0.7,&2\leq x<3\\ 1,&x\geq3 \end{cases}

45. Read probability from that CDF

Find:
P(1<X≤3).P(1<X\leq3).
Using:
F(3)−F(1)F(3)-F(1)
we get:
1−0.2=0.8.1-0.2 = \boxed{0.8}.
This corresponds to:
P(X=2)+P(X=3)=0.5+0.3=0.8.P(X=2)+P(X=3) = 0.5+0.3 = 0.8.

46. Finding PMF from CDF

Suppose CDF jumps:
  • from 00 to 0.20.2 at x=1x=1
  • from 0.20.2 to 0.70.7 at x=2x=2
  • from 0.70.7 to 11 at x=3x=3
Then PMF masses are jump sizes:
P(X=1)=0.2P(X=1)=0.2 P(X=2)=0.5P(X=2)=0.5 P(X=3)=0.3.P(X=3)=0.3.

47. Support of a random variable

The support is roughly the set of values where the random variable can occur with nonzero probability/density.
For the two-head-count example:
supp⁡(X)={0,1,2}\boxed{ \operatorname{supp}(X)=\{0,1,2\} }
For continuous variables, support may be an interval.

48. Indicator random variable

A useful special random variable:
IA={1,A occurs0,A does not occurI_A= \begin{cases} 1,&A\text{ occurs}\\ 0,&A\text{ does not occur} \end{cases}
This converts an event into a numerical random variable.
Then:
P(IA=1)=P(A).P(I_A=1)=P(A).
This becomes very useful later in statistics and probability.

49. Bernoulli-style intuition

If a random variable only takes:
00
or:
1,1,
with:
P(X=1)=pP(X=1)=p
and:
P(X=0)=1−p,P(X=0)=1-p,
it represents a success/failure experiment.
This is the building block for Week 11's binomial distribution.

50. Discrete uniform RV

If a discrete random variable can take nn values and each is equally likely:
P(X=xi)=1n.P(X=x_i)=\frac1n.
Example:
fair die.
This is a discrete uniform distribution.

51. Not every discrete RV is uniform

For two coin tosses and number of heads:
P(X=0)=14P(X=0)=\frac14 P(X=1)=12P(X=1)=\frac12 P(X=2)=14.P(X=2)=\frac14.
So discrete does not imply equal probabilities.

52. Random experiment → RV workflow

text
Random experiment
       ↓
Sample space S
       ↓
Define X(ω)
       ↓
Find possible X values
       ↓
Group outcomes producing same X
       ↓
Compute probabilities
       ↓
PMF / CDF
This workflow solves many Week 9 problems.

53. Full example

Experiment:
Toss three fair coins.
Define:
X=number of heads.X=\text{number of heads}.
Possible values:
0,1,2,3.0,1,2,3.
Total outcomes:
23=8.2^3=8.

54. Construct PMF

Exactly 0 heads:
TTTTTT
so:
P(X=0)=18.P(X=0)=\frac18.
Exactly 1 head:
HTT,THT,TTHHTT,THT,TTH
so:
P(X=1)=38.P(X=1)=\frac38.
Exactly 2 heads:
HHT,HTH,THHHHT,HTH,THH
so:
P(X=2)=38.P(X=2)=\frac38.
Exactly 3 heads:
HHHHHH
so:
P(X=3)=18.P(X=3)=\frac18.

55. Check normalization

18+38+38+18=88=1.\frac18+\frac38+\frac38+\frac18 = \frac88 = 1.
So the PMF is valid.

56. CDF for same example

F(x)={0,x<018,0≤x<112,1≤x<278,2≤x<31,x≥3F(x)= \begin{cases} 0,&x<0\\ \frac18,&0\leq x<1\\ \frac12,&1\leq x<2\\ \frac78,&2\leq x<3\\ 1,&x\geq3 \end{cases}

57. Example CDF query

Find:
P(X≤1).P(X\leq1).
Directly:
F(1)=12.F(1)=\frac12.
Or:
P(X=0)+P(X=1)P(X=0)+P(X=1) =18+38=12.= \frac18+\frac38 = \frac12.

58. Another query

Find:
P(1<X≤3).P(1<X\leq3).
Using CDF:
F(3)−F(1).F(3)-F(1).
Thus:
1−12=12.1-\frac12 = \boxed{\frac12}.
Equivalent values:
X=2,3.X=2,3.
Common mistake: probabilities need not sum to 1 over a subset
For the entire support:
∑xp(x)=1.\sum_xp(x)=1.
But if you sum only selected values, you get the probability of that event, which may be less than 11.
Common mistake: negative PMF
A proposed PMF cannot have:
p(x)<0.p(x)<0.
Even if all values sum to 11, one negative probability makes it invalid.
Both conditions matter.
Common mistake: CDF decreasing
A CDF can stay flat or increase.
It can never decrease.
So:
text
0.2 → 0.5 → 0.4
cannot be a valid CDF.
Common mistake: CDF outside
[0,1][0,1]
A valid CDF cannot be:
−0.1-0.1
or:
1.2.1.2.
Always:
0≤F(x)≤1\boxed{ 0\leq F(x)\leq1 }
Common mistake: CDF ending below 1
As:
x→∞,x\to\infty,
the event:
X≤xX\leq x
eventually includes all possible values.
Therefore:
F(x)→1\boxed{ F(x)\to1 }
Common mistake: confuse
P(X<x)P(X<x) with P(X≤x)P(X\leq x)
For discrete variables, they can differ.
For example:
P(X<2)P(X<2)
excludes:
X=2.X=2.
But:
P(X≤2)P(X\leq2)
includes it.
Their difference is:
P(X=2).P(X=2).
Common mistake: use PMF for continuous data
For continuous random variables:
P(X=x)=0.P(X=x)=0.
So a PMF-style table is not the right representation.
Continuous distributions are handled through density and CDF ideas.
Common mistake: sample space = support
Not necessarily.
Example:
Two coin tosses:
Sample space:
{HH,HT,TH,TT}.\{HH,HT,TH,TT\}.
For:
X=number of heads,X=\text{number of heads},
support:
{0,1,2}.\{0,1,2\}.
They are different sets.
Common mistake: every random variable uses integers
No.
Discrete RVs may take values like:
0.5, 1.5, 3.70.5,\ 1.5,\ 3.7
as long as the set of possible values is countable.
Continuous/discrete classification is not about whether values contain decimals.

68. Probability table completion

Suppose:
xxp(x)p(x)
00.2
1kk
20.5
Then:
0.2+k+0.5=1.0.2+k+0.5=1.
Hence:
k=0.3.\boxed{k=0.3}.
This is a standard Week 9 pattern.

69. Parameterized PMF example

Suppose:
p(x)=k(x+1),x=0,1,2.p(x)=k(x+1), \qquad x=0,1,2.
Then:
k(1)+k(2)+k(3)=1.k(1)+k(2)+k(3)=1.
Thus:
6k=16k=1
and:
k=16\boxed{ k=\frac16 }

70. Probability using parameterized PMF

Then:
P(X≥1)=p(1)+p(2).P(X\geq1) = p(1)+p(2).
So:
=16(2)+16(3)= \frac16(2)+\frac16(3) =56.= \boxed{\frac56}.

71. CDF can answer many event questions

Examples:
P(X≤b)=F(b)P(X\leq b)=F(b) P(X>b)=1−F(b)P(X>b)=1-F(b) P(a<X≤b)=F(b)−F(a)P(a<X\leq b)=F(b)-F(a)
For discrete endpoint variations, carefully account for probability masses at endpoints.

72. Probability greater than a value

P(X>x)P(X>x)
is the complement of:
X≤x.X\leq x.
Therefore:
P(X>x)=1−F(x)\boxed{ P(X>x)=1-F(x) }

73. Continuous case preview

For a continuous RV:
P(a<X≤b)=F(b)−F(a).P(a<X\leq b) = F(b)-F(a).
Because point probabilities are zero, endpoint choices do not matter.
This becomes more important in Week 12.

74. Why Week 9 matters

Random variables let probability theory talk about actual numerical quantities.
Without random variables:
Did the customer buy?
With random variables:
How much did the customer spend?
Without RVs:
Did an arrival happen?
With RVs:
How many arrivals occurred?
This is the bridge from event probability to statistical distributions.

75. Week 9 formula sheet

Random variable

X:S→R\boxed{ X:S\to\mathbb R }

PMF

pX(x)=P(X=x)\boxed{ p_X(x)=P(X=x) }

PMF normalization

∑xpX(x)=1\boxed{ \sum_xp_X(x)=1 }

CDF

FX(x)=P(X≤x)\boxed{ F_X(x)=P(X\leq x) }

Interval from CDF

P(a<X≤b)=F(b)−F(a)\boxed{ P(a<X\leq b)=F(b)-F(a) }

Upper tail

P(X>x)=1−F(x)\boxed{ P(X>x)=1-F(x) }

76. Week 9 conceptual map

text
Experiment
    ↓
Sample space
    ↓
Random variable X
    ↓
Possible numerical values
    ↓
Distribution
    ├── Discrete
    │    ├── PMF
    │    └── CDF
    │
    └── Continuous
         └── CDF
             ↓
         PDF later

77. Weeks 7–9 progression

text
Week 7
Events and probability
        ↓
Week 8
Conditional probability
        ↓
Week 9
Random variables and distributions
The abstraction level is increasing:
outcomes→events→numerical random variables\boxed{ \text{outcomes} \rightarrow \text{events} \rightarrow \text{numerical random variables} }

78. What you should actually master

By the end of Stats 1 Week 9, you should be able to:
  1. Define a random variable.
  2. Distinguish sample-space outcomes from random-variable values.
  3. Construct a random variable from an experiment.
  4. Distinguish discrete and continuous random variables.
  5. Define a PMF.
  6. Check whether a proposed PMF is valid.
  7. Use: ∑xp(x)=1.\sum_xp(x)=1.
  8. Solve for unknown PMF constants.
  9. Compute event probabilities from a PMF.
  10. Define a CDF:
F(x)=P(X≤x).F(x)=P(X\leq x).
  1. Construct a CDF from a discrete PMF.
  2. Recover PMF values from CDF jumps.
  3. Use:
P(a<X≤b)=F(b)−F(a).P(a<X\leq b)=F(b)-F(a).
  1. Know that CDFs are nondecreasing and lie in [0,1][0,1].
  2. Understand that a continuous RV has:
P(X=x)=0.P(X=x)=0.
If those are comfortable, Stats 1 Week 9 is under control.

Quick Recall & Flashcards

Test your active recall before proceeding to practice problems. Format:
Question :: Answer.
  • Week 9 Diagnostic: For a discrete random variable X with PMF P(X=1) = 0.2, P(X=2) = 0.5, P(X=3) = 0.3, what is the value of the CDF F(2)? :: Answer: 0.7 — F(2) = P(X <= 2) = P(X=1) + P(X=2) = 0.2 + 0.5 = 0.7.
  • Week 9 Diagnostic: Which of the following is NOT a property of a valid CDF F(x)? :: Answer: C — A CDF is always non-decreasing and approaches 1 as x -> inf.

Practice & Self-Check

Attempt these diagnostic questions to verify your operational understanding.
1

Question 1

For a discrete random variable X with PMF P(X=1) = 0.2, P(X=2) = 0.5, P(X=3) = 0.3, what is the value of the CDF F(2)?
NAT
3 Unit Assessment
2

Question 2

Which of the following is NOT a property of a valid CDF F(x)?
MCQ
3 Unit Assessment
A. 0 <= F(x) <= 1
B. F(x) is non-decreasing
C. F(x) is strictly decreasing as x -> inf
D. lim_{x -> inf} F(x) = 1

Active Recall Session

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