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ECMT2130 discrete random variables
创建日期：2024-06-12 17:47:41
所属分类：数学mathematics
标签： ECMT2130
discrete random variables

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ECMT2130 – Tutorial 3

Conceptual Questions
1) What does it mean for two events to be independent?
It means an event contains no information about another event. Intuitively, if we are told
something about event A, this does not help us understand anything about event B. In terms of
probability, if two events are independent then P(A|B) = P(A), so the likelihood of observing
A isn’t affected by B. Recall that if two events are independent then their correlation is zero,
but not the other way around (necessarily).
2) Provide two examples of discrete random variables and continuous ones.
Discrete: (1) number of times the daily price of a bond will fall in the next month; and (2) the
number of times a football player received a yellow card in a tournament.
Continuous: (1) the exchange rate between the British Pound (GBP) and the Australian Dollar
(AUD); and (2) the average amount of rain that falls in the blue mountains every month.
3) What is the difference between a probability mass function (PMF) and a probability
density function (PDF)? What is the difference between a PDF and a CDF?
The PMF is suitable for discrete distributions while the PDF is appropriate for continuous
distributions. The CDF represents the likelihood of an accumulation of events happening, and
this is measured normally from minus infinity (or a minimum value assumed by the random
variable) until a point of interest. The PDF is designed to evaluate the chance of a range of
values (or even a particular value, but we know this is not interesting). It is important to note
that many questions we can ask about the probability of a range can be answered by both the
PDF and CDF – it depends on the information you have.
You can find the mathematical definitions in the lecture slides.
4) Would one be able to find a derivative for the CDF of a discrete distribution?
No. The CDF of a discrete distribution often involves “kinks”, making it impossible to find a
derivative for.
5) Assuming share prices are a continuous variable, what is the probability of Woodside
share price to be $40 next month?
Zero. Recall that the probability of a particular value of a continuous random variable is zero.
Mathematically:
Pr( = ) = Pr( ≤ ≤ ) = ∫ ()
= 0
So, if the distribution is continuous, we can write Pr( < ) = Pr⁡( ≤ ).
6) Define expected value, both mathematically and in simple words.
The expected value of a random variable (say, X) is the average of all possible values of X
weighted by the probabilities of observing each of these values. It is the first moment of the
distribution, which is a measure of central tendency of a distribution.
Mathematically:
() = ⁡
{
∑ ∗ Pr⁡( = )
=1
, ⁡⁡⁡
∫ ()
∞
−∞
, ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡
7) What do we mean when we say a normal distribution is fully characterised by its mean
and variance?
We mean that we can fully “draw” the distribution if we know the exact values of the mean
and variance of a random variable. Notation-wise, if a random variable follows a normal
distribution, we write ~(, 2). This implicitly indicates we need and 2 to determine
the exact distribution of the random variable R.
8) What does the kurtosis measure?
The kurtosis measures the “fatness” of the tails of a distribution. Intuitively, it measures the
chance of extreme events happening in relation to the mean of the distribution. You can find
the formula for the kurtosis in the lecture slides.
9) What is the skewness and kurtosis of a normal distribution?
The skewness of a normal distribution is zero (indicating perfect symmetry around the
mean/median/mode), while its kurtosis is 3.
10) What is the difference between error term and residuals in regressions?
The error term () is a population concept, while the residual (̂) is an estimate of what the
error term is. Thus, the residual is a sample concept.
Recall that because we don’t have access to the population, the error term is unobservable. But
we can get an idea of how it behaves (if the OLS assumptions are satisfied, of course) by
observing the residuals.
11) What are the 5 essential assumptions for the OLS estimator to produce high-quality
estimates in large samples?
The assumptions are: (1) linearity in parameters, (2) random sampling, (3) no perfect
collinearity, (4) zero conditional mean and (5) homoskedasticity.
12) What does it mean for an estimator to be unbiased?
An estimator is unbiased if its expected value is equal to the true (unobservable) value we are
trying to estimate. Formally, E(estimator) = true value. In the context of regressions with the
OLS estimator: (̂) = .
There are many intuitive ways of understanding unbiasedness. For example, consider the case
of a simple linear regression between wage (dependent variable) and education (independent
variable). For every sample we have access to, we will obtain a different estimate (but all
produced with the same estimator – the OLS). If we obtain infinitely-many estimates, and take
the average of all of them, then we would observe the true/population relationship between
wage and education, if the estimator is unbiased. If the estimator is biased then it’s likely that
the estimator is “missing” something (in my example, this would likely be the case because,
for instance, the number of years of experience a person has also affects their wage and is at
least partially related to education attainment).
13) What does it mean for an estimator to be consistent?
An estimator is consistent if, as the sample becomes larger and larger (i.e., if we include more
observations into our existing sample), the estimates (̂) get closer and closer to the true value
(). Intuitively, a consistent estimator becomes better at guessing what the true value is the
more information we feed into it. Formally:
lim
→∞
Pr(|̂ − | > ) → 0
The expression above states that as the number of observations (n) gets larger and larger, the
difference between the estimates and the true value being larger than a very, very small
number () gets closer to zero. The closer you want ̂ to be of (i.e., the smaller ) the higher
the number of observations necessary.
Numerical and Computation Questions
1) Consider a random variable Y equal to the sum of the outcomes of tossing three fair coins:
= 1 + 2 + 3
Where 1 is the result of tossing the first coin, 2 the second and 3 the third. Assume the
general outcomes for each tossing follows:
= {
1, ⁡
0, ⁡⁡⁡⁡⁡
, = 1,2,3.
For example, if the second throw results in Heads, then 2 = 1.
a) What possible combination of values can (1, 2, 3) assume in this experiment?
The sample space is S = {(H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T,
H), (T, T, T)}. Therefore, the combination of values is {(1, 1, 1), (1, 1, 0), (1, 0, 1), (1, 0, 0),
(0, 1, 1), (0, 1, 0), (0, 0, 1), (0, 0, 0)}.
This results in possible values for Y to be {0, 1, 2, 3}.
b) Find the distribution function of Y. This involves showing all the possible values for Y and
the chance of observing each value.
As calculated in (a), the possible values for Y are {0, 1, 2, 3}. The probabilities associated
with each of them are 1/8, 3/8, 3/8 and 1/8, respectively.
c) Draw the probability mass function (PMF) and the cumulative mass function (CDF) of the
random variable Y.
d) Find the expected value of Y and Y2.
The expected values are given by:
() = ⁡∑Pr( = ) ∗ =
1
8
∗ 0 +
3
8
∗ 1 +
3
8
∗ 2 +
1
8
∗ 3 = 1.5
(2) = ⁡∑Pr( = ) ∗
2 =
1
8
∗ 02 +
3
8
∗ 12 +
3
8
∗ 22 +
1
8
∗ 32 = 3.0

e) Find the variance and standard deviation of Y.
The variance is given by:
2 = () = ⁡(2) − [()]2 = 3 − (1.52) = 0.75
The standard deviation is given by:
= () = √() = √0.75 ≅ 0.87⁡

f) Find the skewness of Y. Hint: if you wish to calculate the skewness, use the fact that
[()] = ∑Pr( = ) ∗ () applied to () = [ − ()]
3. Then, use the formula for
the skewness from the lecture slides.
Does this skewness level make sense when analysing the PMF of Y?
We don’t really need to calculate the skewness because the distribution is perfectly
symmetric. Another way to see this is that the median (
1+2
2
= 1.5) is the same as the mean,
so the skewness is zero.
But we can use the formula for the skewness if you prefer the long way:
() =
[( − ())
3
]
3

Using the hint, we can calculate [( − ())
3
] by using () = ( − ())
3
, so:
[( − ())
3
] =∑Pr( = ) ∗ ( − ())
3
= 0
Because > 0 then, () = 0.
g) Find the kurtosis of Y. Compare it to the kurtosis of a normal distribution.
The kurtosis is given by:
() =
[( − ())
4
]
4

Using the hint, we can calculate [( − ())
4
] by using () = ( − ())
4
, so:
[( − ())
4
] =∑Pr( = ) ∗ ( − ())
4
≅ 1.31
We can obtain the denominator using the square of the variance:
4 = (2)2 = 0.752 = 0.5625
Therefore:
() =
1.31
0.5625
≅ 2.33
The kurtosis is smaller than that of a normal distribution (3).
2) Consider three return series with expected values and variances as in the table below:
a) Calculate ( + + ) and ( + + ). Assume the returns are independent
of one another.
( + + ) = () + () + () = 9.0%
( + + ) = () + () + () = 5%
2 + 10%2 + 13%2 = 2.94%
b) Consider a portfolio with 20% weight on company A, 45% weight on company B and the
remaining in company C? Assuming independence across the returns, calculate the mean and
standard deviation of the portfolio.
() = (∑
=
) = () + () + () = 3.15%
() = (∑
=
) =
2 ∗ () +
2 ∗ () +
2 ∗ () ≅ 0.42%
() = √() ≅ 6.48%
c) How does your answer change if the weights were 30% for company A, 25% for company
B and the remaining for company C? Which portfolio would prefer to buy and why?
() = (∑
=
) = () + () + () = 3.15%
() = √() ≅ 6.54%
Strictly speaking, without knowing the risk preferences (utility function) for the specific
individual making the choice, we cannot tell which one is best (perhaps the individual likes
risk, so they would prefer the second portfolio). But if the individual is risk-neutral or risk-
averse then the first portfolio is better because it offers the same (expected) reward for a
lower risk (measured by the standard deviation).

d) Assume the same portfolio as in (c) but now with (, ) = 0.03, (, ) =
0.04 and (, ) = −0.01. Which portfolio would you prefer now (b, c or d)?
The returns will not be affected by the covariances, only the variances.
() = (∑
=
) =
=
2 ∗ () +
2 ∗ () +
2 ∗ () ⁡+ 2 ∗ ∗ ∗ (, ) ⁡+ 2 ∗
∗ ∗ (, ) + 2 ∗ ∗ ∗ (, )
() ≅ 13.16%
3) A marketing manager of a leading firm believes that total sales for the firm next year can
be modelled by using a normal distribution with a mean of $2.5 million and a standard
deviation of $300,000. You may want to use Excel to calculate the probabilities.
a) What is the probability that the firm’s sales will exceed $3 million?
Let X be the level of sales in millions. We know that X ~ N(2.5,0.32)
Pr( > 3) = 1 − Pr( ≤ 3) = 4.78%
This can be easily obtained in Excel by using the function “=1-NORM.DIST(3,2.5,0.3,1)”
(without the quotes). The command NORM.DIST calculates the CDF of a function at a
specific point.
b) What is the probability that the firm’s sales will fall within $150,000 of the expected level
of sales?
Pr(2.5 − 0.15 < ≤ 2.5 + 0.15) = Pr⁡( 2.35 < ≤ 2.65) =
= Pr( < 2.65) − Pr( < 2.35) = 38.29%
In Excel, use “=NORM.DIST(2.65,2.5,0.3,1)-NORM.DIST(2.35,2.5,0.3,1)”.
c) In order to cover fixed costs, the firm’s sales must exceed the breakeven level of $1.8
million. What is the probability that the sales will exceed the breakeven level?
Pr( > 1.8) = 1 − Pr( < 1.8) = 99.02%
In Excel, use “=1-NORM.DIST(1.8,2.5,0.3,1)”.
d) Determine the sales level that has only 9% chance of being exceeded next year.
Pr( > ) = 0.09
1 − Pr( < ) = 0.09 ↔ Pr( < ) = 0.91
= 2.9022
In Excel, use “=NORM.INV(0.91,2.5,0.3)”.
4) Patrick wants to estimate the relationship between the average returns and risk (measured
by the standard deviation of returns). He collects data for several share prices over time and
takes the average of them; he does the same for the standard deviation. He obtains one
observation (one value for return and one for variance) for each company, so he is dealing
with a cross section (not a time series).
He applies to OLS estimator to obtain the following results:
̂ = 5.87 + 1.12 ∗
Both returns and standard deviation are measured in percentage. So, = 4 means company
i‘s average return was 4%.
a) Interpret the estimate for the intercept coefficient. What does it suggest?
The intercept indicates the average return for a stock is 5.87% when the standard deviation is
0%. This suggests that a risk-free asset will on average pay around 5.87% as return.
b) Interpret the estimate for the slope coefficient.
For every one percentage point of extra risk (standard error), the model suggests the returns
will increase by 1.12 percentage points, on average.
Recall that percentage points are changes in variables already coined in percentages. Because
both returns and standard deviations are in %, their changes are in percentage points.
c) Plot line of best fit.
I trust you can do this one on your own.
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MOEC0021 STA 106 ENGG 1300A ENGG 1300 OLET1139 STAT 3503 STAT 206 FEM11149 ELEN 4720 COMP20003 SOLA1070 CS574 C11CS LINB29H3 MAST90083 ECON7200 STAT 610 LINB29 DSM-5 COMP3161 EEEN40010 MATH 4431 K353 STA 3386 IERG4300 COMS W4167 ENG5298 AMME 2302 CS3910 COMPSCI 320 STATS 369 GR5242 COSC 328 CSCI435 MGTS1301 INFS2200 COMP3900 CIS 375 Module AE03 COMP3121 CS260 HAM503 MGMT5050 INT3075 COSC 285 GEOL0029 GEGE2X01 STAT1201 MATH1061 MMF1941H STAT 5060 CS 520 COMP 4320 MSIN0104 000T FINS3648 ECMT2160 ECON4003 Phys 129L CSE 142 CS 3130 MGEB01 EECS 340 Finance (20443/20444) ECMM149 COMP5310 EEEN60180 CSC 111 AT2 CSCI 4430 STAB57 MA/CSC 427 MATH322 D100 CSCI544 CITS3004 CI6227 CSE 565 APCO 1P01 852L1 ELEC S411F COMP3207 STATS330 ECON7530 FIT5149 CSI4104 CMPT 371 COMP201 SEHS4696 FI4003 MATH6182 CG1 WS MATH401 MGT 205 QTM 110 COMP 0137 MACE12201 FIT5210 ECE 544 BISM7206 HUDM 6055 ECON3203 STATS 210 MSCI 570 MATH6173 COMPSCI4039 Accounting COM3503 AOE2024 ECON213 PSYC10004 Control Data 100/200A PHIL2617 PHYS3116 MIE250 BCPM0008 ECN 620 QBUS6830 ECSE-415 ECON5102 MGMT20005 MAT 653 ECO 480 COMP 5/600 CSSE 5600 ME 571 7CCMMS61 CS 550 COMP2865 MATH3075 STATS 506 QTM 220 ISYS2120 FIT5106 STAT 153 MKTG3112 ECON2102 ELE2018 COMP3031 MSIN0010 SCC461 MATH237 CS246 CSC3065 CSE 101 ICTWEB411 COMP9313 ECOS 2025 FN620 MATH11111 MATH 3018 KE1068 FINA 6112 C++ MISM 6202 MSIN0105 CSC8631 MECO6935 COMP4121 ACCT2011 DPBS1150 EEL 3701C F19PL OPM 661 CSE 310 CS918 BIOL30001 ACCT3563 BANK3600 MET CS622 MATH367 ACCT5907 Continuing COMP5048 COMP61021 CS3334 GSOE9210 CSCI-570 COMP 5322 HW 2 ALY-6020 IEOR 4706 DATA 200 EBUS504 CGE12412 COMP4108 7SSMM800 ECMM171P MATE50002 CS225 RS 6003 BH2274 ELEC3111 MAT6669 ECON-UA 227 PHIL 1012 7SSMM603 TB028A P6 ECS708 CSC 427 CS6476 CSE 250A A1A2 ITS61504 EECS 376 model HW6 UV0010 ECON 6074 MG4F7 CMP-7030Y INFO3008 BFIP013 ECON 5068 MATH6002 CS3483 PSTAT 131 C31FM CSOR W4246 AAEC 5307 DATA ELEC362 COMSM0047 ECN6006 MA20224 BST351 MATH 323 MATH323 AA2021 COMP3811 COMP2013 IB9X60 INVP001 EMATM0051 COMP220 BLAW1002 E20 TCP8001 ECON61001 EART40005 ECONM1022 MATH 365 MATH 367 ELEC6218 COMP 5812M EDUC 5061M ELEC ENG 1101 equation MATH6174 ST334 C4 EC9570 A33648 EBUS603 ECON47815 stata WFMS0001 PEM102 INF6060 ionically ELECTRICAL MATH 375 MMW 12 MKTG860 ECON60401 ALY6010 SOST20131 Introduction COM3524 MME2 GGR305 JNB 538 EECS101 COMM5030 IEOR E4601 2B CSC336 EN2315 LAW 432 CSC263 MATH2001 OLET1143 CSCB63 SIT772 COMM1170 STAT 441 MS930 INFT3050 GLBH0030 Stat 120B MAST10007 MGEC61 Machine GGR 252H1S ECON7030 Financial Marketing MGMT30019 ECON5120 SDSC 8009 CSC165H1S DSCI 551 3081W ACM41000 ECA 5304 MATH 1064 ECSE444 DATA 604 COMP7105 MIE1615 ECON4004 ENV200H1S EC481 MKT 306 EC1B1 MATH08051 ALY6140 CS260C CSE30 MSIN0226 CS4103 CSC3064 AM16 COMP11212 CSE4101 PSTAT 174 Math 380W CNT 5106C STAT 310 FIN2028 CS5014 STAT 425 COMP 4102A ELEC2140 MANF9544 QRM 9 QRM 8 ECON 457 COMPSCI 4091 MATH 523 DNSC-4211 PEAS 6000 MSIN0180 COMP2119 A2A HT 2022 IB93F0 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ACCOUNTING BUSS1040 CHEM 181 MAT136H1 GY 6183 COMP3170 AGRI90089 PHYSICS 1002 ECON10003 IEOR E4004 ACS61012 MATH 40550 MKTG2113 TELE4653 ENSC1004 ECE 232E MK726 MTRX2700 SHEET 3 CSCA08 COMP30027 Math 2XX3 statistic MMAN3200 BISM7213 ECMM109 ECON 2022 FNCE 20005 GR5067 CSCB09 ME 3017 CS 5100 PHYS 2903 COMS 4771 COMP222 COMP 1046 CSC148H1 ISYS2038 FINS2615 OT5206 COMP2048 A6 MECH5770M MA826/20 CMPT 726 STA457 INFS6071 INFS588 integral ECON 216 IRDR 0017 MA323 PA 15213 PSYC2001 ENV222 Statistics ACCT5930 ACCT12 LAND2121 ME 5405 COMM1180 TELE3113 A7 ECE 568 ACCT2019 INFS1200 ENGG1340 CS 1400 ASST3 FINM7409 SWEN30006 INFO2222 FIN20150 SOSS1000 CSSE7231 FINS5513 COMS3200 CECN 702 ECON4060H MN3515 HW2 PSYA02H3 ECON 2112 COMP90051 CS 1652 MA30059 AF1605 MKTG7501 ECO-6003B COM4521 CTM ECON3350 COMP 3804 ELEC9762 BISM 7255 ST22 MATH20014 MNE3122 COMP343 MATH5165 ELE7029 ENGR30002 SEEM3500 ECON7350 RISK5002 FINM7405 COMP90059 MOS 3320B STAT 431 BISM7221 MATH3014 MTH783P EBA 35302 ACS133 BUS2206 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FN1024 MATH0094 MANG 6554 ECOS3003 ELEC6252 C2511C PHY328 EC1060 ECON32202 COMP4002-E1 MATH3063 EFIMM0124 QBUS2310 ECON4411 FUNDAMENTALS ACCT20001 18YX RPM COMP 2711 QBUS6600 BUSS5221 ST3189 FIT1047 MAT344 EGB375 MATH60013 MATH96011 COMP3557 MAST20029 ECM159 EAE1011 ENSCEN 105 SECTION A STA 141C SOSE 2022 PAM006 COMP0048 M40007 ISOM2500 CMPE 142 A06472 MCC150 ACST 8083 GEOG6087 ECON920 FINM7008 ENVENG 4110 COMPSCI 361 ENGRCEE 21 PHYS 1112 ACF2100 ECMM108 COMPS267F MAT00050M MAST30025 CSIT970 MSCI 224 INT202 PUBL0054 MSDM5004 BMAN 21020 BMAN21020 MGMT1002 MATH11150 LAWS7023 ACTL30002 CIVL3340 MATH1002 ECON 8069 COMPSCI 351 IBUS7302 STAT4528 CHEM1101 STAT3925 BUSN 7008 Comp 3120 ECOS2040 SWEN90010 ENGN6528 ECM158 ECON20022 COMP6452 Calculation CITS1401 S1889112 BUS 351 ACTL30008 FIT2091 ECON6002 IBUS6020 PHI 001 ECON30290 MATH7501 ECON10004 MFIN7017 ISYS90043 LAWS7012 ECON30025 COMP30023 COMM1190 ECOM20001 GSBS6009 STAT 337 EMET8002 ECOM90001 ECOM30001 FINC 2012 BUSINESS ENGCV512 ECO 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