Questions and Answers of Data Science

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Data Science
Student’s Name
Affiliation
Course
Instructor
Date

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Question 1
A random variable V has a density fv(v) =Kv2 exp(-b/v2 )
a) What is the distribution of a new random variable, Y=1/2 mV2
The first step will entail differentiation of the function fv(v) =Kv2 exp(-b/v2 ) w.r.t V
d (f ( v ))
d(v ) =2KV* exp (-b/v2) + KV2*-2e-bv-3
The distribution for the new variable is given by
F(y)= 2KV* exp (-b/v2) + KV2*-2e-bv-3
b) in order to get the mean of y one has to differentiate y with respect to V and get the
d (f ( y ) )
d( y) = MV
The mean of Y=MV
Question 2
Que. A
> # the expected value of a binomial distribution is given by np
> #n is the number of observations
> #p is the number of success
> #p=6/10
> #n=10
> #let expected value be represented by M
> p<-6/10
> n<-10
> M<-p*n
> M

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[1] 6
> #this means that the mean is 6
Rcode:
Que B
The posterior distribution is given by
P(Ɵ) ={ p=3/5 for x=1,2…..10
The estimated value for (Ɵ) is =Ɵ/6 where 6 is the mean of the distribution
Que. C
> p<-c(0,0.15,0.25,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1)
> p1<-c(0.025,0.05,0.8,0.06,0.03,0.01,0.008,0.006,0.005,0.004,0.002)
> plot(p~pl)

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Rcode
Question 3
The MLE (Ɵ) ( L Ɵ ) = Ɵ X1 Ɵ-1 *Ɵ X2 Ɵ-1 *Ɵ X3 Ɵ-1 *Ɵ Xn Ɵ-1
= Ɵ(∑
x=1
n
XƟ−1
= Ɵ(
∑
x=1
x=n
XƟ
∑
x=1
x=n
X
)
Substituting ∑
x=1
x=n
X with P we have that
MLE (Ɵ) ( L Ɵ )= Ɵ* PƟ
P
Differentiating both sides w.r.t Ɵ we have that

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d (L (Ɵ ))
d (Ɵ) = 1* ƟP^Ɵ-2
Ɵ=¿ ƟP^Ɵ-2
Dividing both sides by Ɵ
1=P^Ɵ/P^2
P^2= P^Ɵ
2logP= Ɵ log P
Dividing both sides by log P
Ɵ=2
b) the MLE (Ɵ) ( L Ɵ )= Ɵ* PƟ
P
P is given by ∑
x=1
x=n
X
As the value of X increases it means that since Ɵ is fixed, the variance will decrease.
Question 4
a) The poison probability distribution is given by:
P(X:u)= ( e−u )∗(ux)
x ! , where x=1,2,3…n where u is the mean and x is the observed variable
P(Xi=x/X1=1) = P ¿ ¿ = e−i
e 1 . since the base for the denominator is equal to the base for the
numerator, then the equation is simplified into
P(Xi=x/X1=1) = e−i−1
b) The MLE is computed by getting the product of observable outcomes (Smirnov, 2011).
In this case one should multiply the function for all values of x in order to get a joint
distribution function

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L(U) = ( e−u )∗(ux1 )
x 1! * ( e−u )∗(ux 2)
x 2 ! * ( e−u )∗(ux3 )
x 3 ! *… ( e−u )∗(uxi )
xi! (A’Hearn, 2004)
=e-nu(∑
i=1
n
ux / x !)
Taking differential from both sides the results are:
d (L (u ))
d (u) = -u* e-nu *(∑
i=1
n
ux / x !)
It therefore follows that the first part of the differentiation gives 1-e-u while the second
part gives ∑ ixi /n. Combining the two parts, the results for MLE becomes
Ǔ= 1-e-u *∑ ixi /n.
Substituting ∑ ixi /n with ẍ then the results becomes
Ǔ=1-e-u * ẍ
c) When ẍ= 3.2,the value for Ǔ is given by
Ǔ= 3.2* (1-e- Ǔ )
Ǔ
3.2 + e- Ǔ =1
Ǔ +3.2 e−Ǔ
3.2 =1
The value of u=0 for the equation above to be satisfied.
If zero replaces u the equation becomes
0+3.2 e−0
3.2 =1
Since any number raise to the power of zero=1, then
0+3.2
3.2 =1 which confirms that the value of Ǔ =0

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Bibliography
A’Hearn, B., (2004). A restricted maximum likelihood estimator for truncated height
samples. Economics & Human Biology, 2(1), pp.5-19. (A’Hearn, 2004)
Smirnov, O., (2011). Maximum Likelihood Estimator for Multivariate Binary Response
Models. SSRN Electronic Journal,. (Smirnov, 2011)