# Finance Analyst Interview Questions

Finance analyst interview questions shared by candidates

## Top Interview Questions

### Finance Analyst at Morgan Stanley was asked...

3. Stock A has a volatility of 20%, B has a volatility of 30%, and their correlation coefficient is 50%. They have the same expect returns. Suppose we can buy x share of A, and 1-x share of B. Then what the x should be for us to invest to make our portfolio has the minimum volatility? 1 AnswerThe vols add in quadrature, with the caveat that the interference term has the correlation factor in there. So: sig(port)^2 = x^2*sig(A)^2+(1-x)^2*sig(B)^2+2*x*(1-x)*sig(A)*sig(B)*rho(AB) Take the partial with respect to x and set that equal to zero. Solve for x and get: x = 6/7 |

Why are dividends not part of the Income Statement? 1 Answerbecause dividends are shareholder income and not firm income. the income statement only contains firm income. dividends are listed under shareholder equity. |

### Financial Analyst at Houlihan Lokey was asked...

decompose the CAPM model and discuss limitations 3 Answerssmall company premium, international premium, etc The CAPM which is E(r) = Rf + Beta *(Rm A model that describes the relationship between risk and expected return and that is used in the pricing of risky securities. Capital Asset Pricing Model (CAPM) The general idea behind CAPM is that investors need to be compensated in two ways: time value of money and risk. The time value of money is represented by the risk-free (rf) rate in the formula and compensates the investors for placing money in any investment over a period of time. The other half of the formula represents risk and calculates the amount of compensation the investor needs for taking on additional risk. This is calculated by taking a risk measure (beta) that compares the returns of the asset to the market over a period of time and to the market premium (Rm-rf). Capital Asset Pricing Model (CAPM) consists of E(r) = Rf + B(Rm-Rf) Rf1 = Risk free rate which usually consists of Beta = company's stock performance against stock market Rm = Expected return from market CAPM ASSUMPTIONS Investors hold diversified portfolios This assumption means that investors will only require a return for the systematic risk of their portfolios, since unsystematic risk has been removed and can be ignored. Single-period transaction horizon A standardised holding period is assumed by the CAPM in order to make comparable the returns on different securities. A return over six months, for example, cannot be compared to a return over 12 months. A holding period of one year is usually used. Investors can borrow and lend at the risk-free rate of return This is an assumption made by portfolio theory, from which the CAPM was developed, and provides a minimum level of return required by investors. |

How would you handle an irrate internal customer during a business meeting? 2 AnswersGood indication of behaviors that are common in this organization. No job is worth total humiliation. I would definetly keep my cool and act very professional and say... I understand your concerns. Please send me your issues in an email and I will make sure I set up a conference call so we can discuss them in further detail. |

### Financial Analyst at Wells Fargo was asked...

How did I handle finding a discrepancy of details. 1 AnswerI reported to my leader and we resolved the discrepancy through analysis and investigation. |

What is the probability of throwing 11 and over with 2 dices 14 Answers1/12 If you are rolling two dice the chances of getting an 11 and OVER is 2/12= 1/6. You can roll a 6 and a 5 to equal 11 and you can roll a 6 and a 6 and get 12 which is OVER 11, therefore the answer should be 1/6. The answer is 1/12 because there are two ways of rolling 11, rolling a 5 and a 6 or rolling a 6 and a 5 (for probability purposes there are indeed two different results). There is only 1 way to roll over 11 (two 6's). That means that out of 36 possible combinations, 3 qualify as being equal to or greater than 11, which means that the probability is 3/36, which reduces to 1/12. Show More Responses All possible results when throwing 2 dices ( 2-3 combination IS THE SAME as 3-2, you know it if you ever played dice, I don't know what are the probability purposes Casey is talking about ): (1,1) (1,2) (2,2) (1,3) (2,3) (3,3) (1,4) (2,4) (3,4) (4,4) (1,5) (2,5) (3,5) (4,5) (5,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) so there are 21 combinations, only (5,6) and (6,6) are >=11, so the answer is: 2/21=9.52% Casey's right, 3/36. Peter, it isn't an ordered problem but there are two rolls out of the 36 possible that can be 11 (6 on die a, 5 on die b, and vice versa). Add that to the one way you can roll boxcars (6 on die a, 6 on b), totally of 3 winning instances out of 36 possible. OK, you are right Casey...I made myself believe that what matters is the final sum, so unless they come up with dice that have 12 sides...I'm wrong :) Thanks Grant I think the question is oriented to find out if you are actually listening. You can not throw 11 AND over. You can throw 11 OR over. So, the answer is 0. Impossible. "Dices"? Did you just say "dices"? Excuse me for answering your question with a question, but what was the highest grade of grammar school education you recall passing? If I got this job, is there any possibility, that is, any chance higher than zero, that you would be my boss? How did you get to your current position in this organization? Can I go now? I'm with Jeff and Carlos- "dice" is the plural of "die". Geez. However, if the question actually is "what is the probability of getting 11 or greater on a single throw with a pair of dice," then the answer is 50%. It either will or will not be 11 or greater. 1) Same as Jeff - "dices" makes no sense 2) How many sides are there on each die? Don't assume that all dice have six sides. 5/36 The answer is there are 6 sides on each die, therefore, 6 times 6 is 36, giving us all the possible outcomes. You can roll an 11 one of two ways a, 5 and a 6 or a 6 and a 5. You can roll a 12 only one way. This gives us 3 possible outcomes of 11 or over out of 36. Which reduces to 1/12. A single die has six possible sides it can land on. That means two die have 36 possible outcomes, but only 18 unique. The probability that the two rolls die sum to 11 is 1/18 He answer is 1/18, not 1/12. Show More Responses *the answer is 1/18 for the sum of an 11 and 1/9 for 11 and over. The correct answer is 1/9. (5,6)-(6,5)-(6,6)-(6,6) 4/36=1/9 |

### Financial Analyst at Two Sigma was asked...

You are to write a method getMissingLetters, which takes a String, sentence, and returns all the letters it is missing (which prevent it from being a pangram) 9 AnswersFirst you have to know that a pangram is a string, sentence, that uses every letter of the alphabet at least once. Comes from the greek, Pan (every) and gramma (letter). If you didn't know that, couldn't figure out the derivation of the word, or weren't told it in the interview - you might be screwed. Otherwise, getMissingLetters should allocate a string (or pointer to an array of chars if C/C++) in which to return the result, a static string which basically just holds the letters of the alphabet, and an array of 27 single ints to hold the count of the number of times a particular letter appears in the sentence. Then create two loops, the outer loop will move through the alphabet (index 0 to 26) on the static string/array, the inner loop will index into (or moving a pointer along) the string holding the input sentence (exiting if you've reached the end of the string (in C/C++ this is '\0' the null char). Within that inner loop you want to check the current letter being pointed to in the alphabet against the current letter being pointed to in the sentence (be careful if writing C/C++ code as to how the sentence is stored in memory). If the value of the characters being pointed to for each array are equal, then increment the value of the integer at the array position with the same index as the current letter being counted. After each inner loop completes, the outer loop counter will move the index into each array for you (or you can increment the pointers to the alphabet and alphabet count arrays). After these two arrays complete, you need to finalize the results. To do this you need one last loop to cycle once more through the alphabet string and alphabet counter. Within this loop you will check the value of the letter count. If it is greater than 0 you move to the next letter, if it is 0, then you want to copy the value of the letter into the current character pointed to in the result string (which should have at least 26 characters assuming a sentence is at least 1 character). Once this loop completes just return the return string (or a pointer to it) In C++ you might be able to utilize some of the STL Classes to do all of this, but if you do it in straight C with lots of pointer arithmetic it will execute much more quickly. Strong no hire Assuming ASCII string: void PrintMissingCharacters(char *InputString) { UINT32 foundChars = 0xFC000000; char* stringChar = InputString; int bitIndex; while (stringChar != '\0') { if (stringChar >= 'a' && stringChar = 'A' && stringChar <= 'Z') { bitIndex = stringChar - 'A'; BitSet(&foundChars, bitIndex); } } for (int i = 0; i < 26; i++) { if (!BitTest(&foundChars, i)) { printf("%c, ", ('a' + i)); } } Show More Responses Using Haskell to keep things terse: include Data.Char getMissingLetters sentence = filter (`notElem` map toLower sentence) ['a'..'z'] or for the more internationally inclined, who might be less biased towards English. generalizedGetMissingLetters sentence alphabet = filter (`notElem` map toLower sentence) alphabet Main> getMissingLetters "hello" "abcdfgijkmnpqrstuvwxyz" #!/usr/bin/perl $s='The good brown fox, hi.'; print join '', map {$s!~/$_/?$_:''} a..z; #!/usr/bin/python """\ Example solution using Python """ import string def getMissingLetters(sentence): """Use sets to find unused characters from the alphabet """ return ''.join(set(string.lowercase)-set(sentence.lower())) Why were you asked this question for a financial analyst interview? #include #include using namespace std; string CheckPangrams(const string str) { string s = str; string result = ""; int hist[26] = {}; for (int i=0; i import java.util.*; public class PangramChecker { private final Character[] charSet = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'}; private Set alphabetSet = new TreeSet(Arrays.asList(charSet)); private Set inputSet = Collections.emptySet(); public PangramChecker(String s) { if (s == null) { s = ""; } inputSet = new HashSet(s.toCharArray().length); for (Character c : s.toCharArray()) { inputSet.add(c); } } public Set check() { alphabetSet.removeAll(inputSet); System.out.printf("\n Missing chars are: " + alphabetSet); return alphabetSet; } public static void main(String[] args) { Scanner scanner = new Scanner(System.in); String s = scanner.nextLine(); PangramChecker pangramChecker = new PangramChecker(s.toLowerCase()); pangramChecker.check(); } } import java.util.Arrays; import java.util.List; import java.util.Set; import java.util.TreeSet; import static org.junit.Assert.*; public class PangramCheckerTest { @Before public void setUp() throws Exception { } @Test public void testCheck(){ String s = "Hello World!"; PangramChecker pangramChecker = new PangramChecker(s.toLowerCase()); Set result = pangramChecker.check(); Character[] charArray = {'a','b', 'c', 'f', 'g', 'i', 'j', 'k', 'm', 'n', 'p', 'q', 's', 't', 'u', 'v', 'x', 'y', 'z'}; List l = Arrays.asList(charArray); Set expectedOutput = new TreeSet(l); assertTrue(result.equals(expectedOutput)); } } although this does not check the char boundary "a" to "z" |

### Financial Analyst at J.P. Morgan was asked...

You have a bat and a ball. The bat is 1 dollar more than the ball and they both add up to 1.50. How much are each? 10 Answers1.25 and 0.25 Why not 1.50 and 0.00 if question does not place any restrictions on values for bat and ball. Why not $1.00 for bat and $.50 for the ball Show More Responses 1.25 for the bat and .25 for the ball... You could not value the bat at 1.50 or 1.00, because they are not equal to 1 dollar more than the ball. 1.25 for the bat and .25 for the ball... You could not value the bat at 1.50 or 1.00, because they are not equal to 1 dollar more than the ball. Are you guys serious? The ball is $1 more!!! 1.50 and 0 would be $1.50 more!! Do not apply. I knew there would be at least a couple who said 1 and .50 ... lol. It's 1.25 and .25 . The price of the bat has to be a dollar more than the price of the bat . If the ball is .50 cents then the bat would be 1.50 . Meaning that in total it would be 2 dollars . Now as for the bat being 1.50 and the ball being 0.00 . The bat has to be a dollar more than the ball. A dollar more than 0.00 is 1.00 . So the ball and the bat together would only be 1.00. When the ball is .25 , the bat would be priced at 1.25. When you add the ball and the bat together (1.25 + .25) the total is 1.50 . ball costs x and bat costs x+1.the adds up x+(x+1)=1.5 so 2x=0.5 and x=0.25 so ball costs 0.25 and bat cost x+1 which is 1+0.25=1.25 $1:25 for the bat, and $0.25 for the ball. |

### Finance Operations Analyst at Google was asked...

Pretend 1% of the population has a disease. You have a test that determines if you have that disease, but it's only 80% accurate and 20% of the time you get a false positive, how likely is it you have the disease. 12 Answers1% divide by reciprocal, multiply by .8% = 1% [1] Fact: 1% of the population has the disease (given) [2] Data: Test is only 80% accurate, and 20% inaccuarate (given): Assume, Population = 10,000 people 1% have the disease = 100 people 99% do not have the disease = 9,900 people Of the 1% who have the disease 80% tested +ve = 80 Of the 99% who don't have the disease 20 tested +ve = 1980 [3] Question: How likely is it that you have the disease? To identify that you have the disease you have to test +ve and actually have the disease = 80 / (80+1980) = 80 / 2060 = 3.88% Show More Responses Its easier if you draw this out in quadrants, Accuracy = 80% = P(Test shows true and having disease) + P (Test showing false and not having disease) Also, we know false positive rate = P(Test shows true but not having disease) = 20% That leaves us with the fact that P(Test being false and person having disease) = 0% Hence likelihood of having the disease is 1% You don't need the test. The answer is given... 1% of the population has the disease... Answer: 1% Very similar question, with a step by step walk through to the solution: http://www.mathsisfun.com/data/probability-false-negatives-positives.html Answer is 3.88% 1% Given: 1. 1% of the population has a disease 2. A test exists to identify disease carriers The questions makes no statement regarding whether or not you have taken the test, the second piece of data has no relevance. You have a 1% chance If you have test positive, how likely is it you have the disease = All sick with positive result / All positive (sick and healthy). All sick with positive result = 1% = (1% x 80%) accurate positive result + (1% x 20%) non accurate but non false positive result. All positive = all sick with positive result + all healthy with positive resultall = 1% + (99% x 20%) = 1% + 19,8% = 20.8% ANSWER = 1% / 20.8% = 5,050505% I thinks this question is ambiguous. If it meant the possibility that you have the disease, then 1%; If it meant the possibility that your test result shows you have the disease, then 1%*80%+99%*20% Sample: 1,000 X=10 P=.8 If the test is only 80% accurate then 8 of the 10 infected will be positive and 2 will walk around unbeknownst, but 20% of the non-infected population will be falsely identified, 1000-10=990 (non-infected) 990*.2=198 (false positive) + 8 (true positive)= 296 (total positive tests) Of this population only 8/296=2.7% were correctly identified. This is the chance you have the disease I think the question is not complete. It should read: "Pretend 1% of the population has a disease. You have a test that determines if you have that disease, but it's only 80% accurate and 20% of the time you get a false positive, how likely is it you have the disease "if you got tested positive. " Use four quadrants method and imagine total number of people = 1000 + Tested -Tested ------------------- ----------------------------------------------------------------- D | 80% of 10 = 8 | 10 - 2 = 8 | 10 ND | 20 % of 990 = 199 | 990 - 199 = 791 | 990 ---------------------------------------------------------------------------------------- | 207 | | 1000 So if one is tested +, they are in first column. Total of which is 207. However only 8 of them actually have disease. So probability that test came out + and the person has disease will be 8/207% = 3.865 ~ 3.7% 1%. The first sentence gives you all the info you need. |

### Financial Analyst at Goldman Sachs was asked...

You are not from a target school, what are you doing here? 5 AnswersSell yourself Talk to them about extracurricular activities and your belief that work ethic is more important than what fancy school you went to. Elaborate as needed. make my school a target Show More Responses To show you why you need to make my school a target. Colorado State Rams! To prove target school is not something your company will be proud of , I am. |

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