Random Variables & Distribution Functions MCQ Quiz in मराठी - Objective Question with Answer for Random Variables & Distribution Functions - मोफत PDF डाउनलोड करा

Explanation -

(i). The prior distribution on ${\phi }^2$ is an inverse gamma distribution

This statement is correct if β > 0. The inverse gamma distribution takes the form $\pi (x)\propto x^{-\alpha }{e}^{-b/x},$ which in this case would translate to $\rho ({\phi }^2)\propto ({\phi }^2{)}^{-(\beta +1)},$ indicating that an inverse gamma distribution is applicable in this case.

(ii). The posterior distribution of ${\phi }^2$ after observing $Y_1,...,Y_n$ is proportional to $(1/{\phi }^2{)}^{n/2+\beta }\times e^{(-nY{̄}^2/2{\phi }^2)}$

This statement is also correct. Bringing together the likelihood function of the sample $Y_1,...,Y_n$ and the prior distribution of ${\phi }^2$ yields the posterior distribution of this form.

(iii). The joint prior distribution of (μ, ${\phi }^2$) is a normal-inverse gamma distribution when μ has a normal prior distribution with mean 0 and variance ${\phi }^2$

This statement is correct. The situation described is essentially the definition of the normal-inverse gamma distribution.

(iv). The posterior mean of ${\phi }^2$ is $(nY{̄}^2+\beta )/((n/2)+\beta -1)$when β > 1/2 and n is large

This statement is not correct. The posterior distribution of ${\phi }^2$ is an inverse gamma distribution and its mean (if it exists) does not hold the specified form. When β <= 1, the posterior mean does not exist.

Hence the option (i), (ii) and (iii) are correct.

Concept:

(i) Let X be a continuous random variable follows uniformly distribution U(a, b) then probability density function

f(x) = $\frac{1}{b-a}$

E(X) = $\frac{a+b}{2}$ and

E(α(x)) = ${\int }_a^b\alpha (x)f(x)dx$

(ii) Two random variable X and Y are non correlated if Cov(X, Y) = 0

Explanation:

A continuous random variable X follow Uniform (−1, 1)

so f(x) = $\frac{1}{2}$, E(X) = $\frac{1-1}{2}$ = 0

Also Y = X2

So Y and X are dependent

Now,

Cov(X, Y) = E(XY) - E(X)E(Y)

                = E(XY) (as E(X) = 0)

               = E(X³) (since Y = X2)

               = ${\int }_{-1}^1{x}^3.\frac{1}{2}dx$ = 0 (as x³ is odd function)

Cov(X, Y) = 0

So X and Y are non-correlated

Therefor option (4) is TRUE and (1), (2), (3) are NOT TRUE

So option (1), (2), (3) is correct

Concept:

The cumulative hazard function H(t) is defined by

H(t) = - $\frac{f}{R}$ where R = 1- F, f be probability density function, R be the survival function and F be the cumulative density function

Properties of cumulative hazard function H(t):

(i) H(t) ≥ 0

(ii) H(t) is either decreasing of increasing or constant function

(iii) H(t) is unbounded function and ​${\displaystyle \underset{\mathrm{t}\to \infty }{lim}}$​​​ H(t) = ∞

Explanation:

By the direct properties of H(t), (1), (2) and (4) are correct

We know that h(t) is unbounded and H(0) = 1 but it's not necessary that H(1) = 1

Option (3) is false

Concept:

A is said to be consistent for mean μ if E(A) = μ as n → ∞ and Var(A) = 0 as n → ∞

Explanation:

{Xi: 1 ≤ i ≤ 2n} be independently and identically distributed normal random variables with mean μ and variance 1, and independent of a standard Cauchy random variable W

So X_i ∼ N(μ, 1) and W ∼ Cauchy(0, 1)

E(Xi) = μ, Var(Xi) = 1, E(W) = 0, Var(W) = 1

(1): ${\mathrm{n}}^{-1}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}}$

E(${\mathrm{n}}^{-1}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}}$) = $\frac{1}{n}{\displaystyle \sum _{i=1}^{n}E(X_i)}$ = $\frac{1}{n}.n\mu$ = μ as n → ∞

Var(${\mathrm{n}}^{-1}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}}$) = $\frac{1}{n^2}{\displaystyle \sum _{i=1}^{n}Var(X_i)}$ = $\frac{1}{n^2}.n$ = $\frac{1}{n}$ = 0 as n → ∞

Therefore ${\mathrm{n}}^{-1}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}}$ is consistent.

Option (1) is correct

(2): ${\mathrm{n}}^{-1}{\displaystyle \sum _{\mathrm{i}=1}^{2\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}}$

E(${\mathrm{n}}^{-1}{\displaystyle \sum _{\mathrm{i}=1}^{2\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}}$) = $\frac{1}{n}{\displaystyle \sum _{i=1}^{2n}E(X_i)}$ = $\frac{1}{n}.2n\mu$ = 2μ as n → ∞

Therefore ${\mathrm{n}}^{-1}{\displaystyle \sum _{\mathrm{i}=1}^{2\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}}$ is not consistent.

Option (2) is not correct

(3): ${\mathrm{n}}^{-1}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{2\mathrm{i}-1}}$

E(${\mathrm{n}}^{-1}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{2\mathrm{i}-1}}$) = $\frac{1}{n}{\displaystyle \sum _{i=1}^{n}E(X_{2i-1})}$ = $\frac{1}{n}.n\mu$ = μ as n → ∞

Var(${\mathrm{n}}^{-1}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{2\mathrm{i}-1}}$) = $\frac{1}{n^2}{\displaystyle \sum _{i=1}^{n}Var(X_{2i-1})}$ = $\frac{1}{n^2}.n$ = $\frac{1}{n}$ = 0 as n → ∞

Therefore ${\mathrm{n}}^{-1}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{2\mathrm{i}-1}}$ is consistent.

Option (3) is correct

(4): ${\mathrm{n}}^{-1}\left({\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}+\mathrm{W}}\right)$

E(${\mathrm{n}}^{-1}\left({\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}+\mathrm{W}}\right)$) = $\frac{1}{\mathrm{n}}\left({\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{E}({\mathrm{X}}_{\mathrm{i}})+\mathrm{E}(\mathrm{W})}\right)$ = $\frac{1}{n}(n\mu +0)$ = μ as n → ∞

Var(${\mathrm{n}}^{-1}\left({\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}+\mathrm{W}}\right)$) = $\frac{1}{{\mathrm{n}}^2}\left({\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{V}\mathrm{a}\mathrm{r}({\mathrm{X}}_{\mathrm{i}})+\mathrm{V}\mathrm{a}\mathrm{r}(\mathrm{W})}\right)$ = $\frac{1}{n^2}(n+1)$ = $\frac{1}{n}+\frac{1}{n^2}$ = 0 as n → ∞

Therefore ${\mathrm{n}}^{-1}\left({\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}+\mathrm{W}}\right)$ is consistent.

Option (4) is correct.

Concept:

If f₁ and f2 are two probability density functions then αf1 + βf2 is also probability density function if α + β = 1 

Explanation:

f1 and f2 denote the density functions of X1 and X2,

and f(y) = 0.4 f​1(y) + qf2(y) is probability density function then

Y = 0.4X1 + qX2 is a random varibale and 

0.4 + q = 1 ⇒ q = 0.6

Option (1), (4) are correct

Given X1 follows a gamma distribution with mean 10 and variance 10, and X2 ∼ N(3, 4)

So E[X1] = 10, Var[X1] = 10, E[X2] = 3, Var[X2] = 4 

E[Y] = 0.4E[X1] + qE[X2] = 0.4 × 10 + 0.6 × 3 = 5.8

Option (2) is correct

Var[Y] = (0.4)²Var[X1] + q²E[X2] = (0.4)² × 10 + (0.6)² × 4 = 3.04

Option (3) is correct

Given:-

X1, X2, ... are i.i.d. random variables having a χ2-distribution with 5 degrees of freedom.

Concept Used:-

The limiting distribution of the given expression can be found using the central limit theorem.

The central limit theorem states that the sum of many independent and identically distributed random variables, properly normalized, converges in distribution to a normal distribution.

Explanation:-

Here, we have n i.i.d. random variables with a χ²-distribution with 5 degrees of freedom.

The mean of each χ²-distributed variable is 5 and the variance is,

2 × 5 = 10

Therefore, the mean of the sum of n such variables is n5, and the variance is,

⇒ variance = (n × 10)

We can normalize the expression by subtracting the mean and dividing by the standard deviation. That is,

$\Rightarrow a[{\displaystyle \frac{(X_1+X_2+...+X_n-5n)}{\sqrt{n\times 10}}}]$

$=({\displaystyle \frac{a}{\sqrt{10}}})[{\displaystyle \frac{(X_1+X_2+...+X_n-5n)}{n}}]\sqrt{n}$

The term in the brackets on the right-hand side is the sum of n i.i.d. random variables with a mean of 0 and a variance of 1/2.

Therefore, by the CLT, this term converges in distribution to a standard normal distribution as n goes to infinity.

The overall expression converges in distribution to a normal distribution with mean zero and variance a²/10.

So, the limiting distribution of $a\left(\frac{X_1+\cdots +X_n-5n}{\sqrt{n}}\right)$ is ​​the standard normal distribution for an appropriate value of a.

Hence, the correct option is 3.

Solution:  

we use the properties of the cumulative distribution function (CDF).

Step 1: Calculate $P(1<X\le 2)$

Using the CDF properties, for $a<X\le b$ :

$P(a<X\le b)=F(b)-F(a)$.

Here, a = 1 and b = 2 .

For 0 ≤ x < 3 , the CDF is given by $F(x)=\frac{x+2}{5}$ .

Substituting these values:

$F(2)=\frac{2+2}{5}=\frac{4}{5},F(1)=\frac{1+2}{5}=\frac{3}{5}.$

Thus:

$P(1<X\le 2)=F(2)-F(1)=\frac{4}{5}-\frac{3}{5}=\frac{1}{5}.$

Step 2: Calculate P(X = 0)

The probability P(X = 0) corresponds to the probability mass at X = 0 .

Since the given random variable is continuous, P(X = 0) = 0 .

Step 3: Add the probabilities

Combining the results:

$P(1<X\le 2)+P(X=0)=\frac{1}{5}+0=\frac{1}{5}.$

Hence the correct option is (1)

Concept:

1. Law of Large Numbers (LLN):

The Law of Large Numbers states that as the sample size n increases, the sample average (or sum) of i.i.d.

random variables converges to the expected value of the variable. For example, in Option 3, $\frac{1}{n}\sum _{i=1}^n{X}_i^2$ 

converges to the expected value $E(X_1^2)=1$, because $X_1$ has variance 1.

2. Central Limit Theorem (CLT):

The Central Limit Theorem tells us that the sum (or scaled average) of i.i.d. random variables with finite mean and

variance converges in distribution to a normal distribution as $n\to \mathrm{\infty }$ . For instance, in Option 4,  $\frac{\sum _{i=1}^n{X}_i}{\sqrt{n}}$ behaves like a

standard normal random variable as  $n\to \mathrm{\infty }$ , converging to N(0, 1) .

3. Probability Limits:

For certain random variables, their distribution converges to a fixed probability value. In Options 1 and 4,

the probability of the standardized sum being less than or equal to 0 converges to $\frac{1}{2}$, which is the probability

that a standard normal variable is less than or equal to 0.

Explanation:

Option 1: This expression involves the ratio of two terms: $\sqrt{n}\sum _{i=1}^n{X}_{i}and\sum _{i=1}^n{X}_i^2$.

The numerator,$\sqrt{n}\sum _{i=1}^n{X}_i$, grows like $O(\sqrt{n})$ by the Central Limit Theorem since the sum of i.i.d. random

variables with mean zero and variance 1 tends to have a normal distribution.

The denominator, $\sum _{i=1}^n{X}_i^2$ , grows like O(n) because it is the sum of the squares of i.i.d. random variables with variance 1.

As $n\to \mathrm{\infty }$, the ratio $\frac{\sqrt{n}\sum _{i=1}^n{X}_i}{\sum _{i=1}^n{X}_i^2}$ tends to zero, so the probability converges to the probability of the random variable being

less than or equal to 0. Hence, Option 1 is true, and the limit will be 1/2 because this becomes a standard normal random variable under large n .

Option 2: The numerator $\sum _{i=1}^n{X}_i$ behaves like $O(\sqrt{n})$ , and the denominator $\sum _{i=1}^n{X}_i^2$ behaves like O(n) . Therefore, the ratio $\frac{\sum _{i=1}^n{X}_i}{\sum _{i=1}^n{X}_i^2}$ behaves

like $O(1/\sqrt{n})$ and converges to 0 as $n\to \mathrm{\infty }$ . Thus, Option 2 is true.

Option 3: By the Law of Large Numbers (LLN), the average of the squares of i.i.d. random variables converges to the

expected value of $X_1^2$, which is 1, as $n\to \mathrm{\infty }$. Thus, Option 3 is true.

Option 4: 
$$
By the Central Limit Theorem (CLT), $\frac{\sum _{i=1}^n{X}_i}{\sqrt{n}}$ converges in distribution to a standard normal random variable N(0, 1) .

The probability that a standard normal variable is less than or equal to 0 is $P(Z\le 0)=\frac{1}{2}$. Hence, Option 4 is true.
 

All four options are correct.

Concept:

The sample mean gives an average of the data points in a sample, and the sample variance

measures the spread of the data.

The normal distribution is a continuous probability distribution that is symmetric around the mean,

with most values clustering around the center.

Explanation:

Option 1: The mean of $\overline{X}$ is 2 and the mean of $\overline{Y}$ is -2 . Thus, the mean of $\overline{X}+\overline{Y}$ is
 

$E[\overline{X}+\overline{Y}]=2+(-2)=0$

The variances add for independent random variables:

  $\text{Var}(\overline{X})=\frac{{\sigma }^2}{n}=\frac{4}{12}=\frac{1}{3},\text{Var}(\overline{Y})=\frac{5}{15}=\frac{1}{3}$

Therefore, the variance of $\overline{X}+\overline{Y}$ is $\text{Var}(\overline{X}+\overline{Y})=\text{Var}(\overline{X})+\text{Var}(\overline{Y})=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$

Thus, $\overline{X}+\overline{Y}$ is distributed as $N(0,\frac{2}{3})$ . This statement is true.

Option 2: $S_1^2$ and $S_2^2$  are independent chi-squared random variables scaled by their respective degrees of freedom.

The degrees of freedom for $S_1^2$ is 11 and for  $S_2^2$ is 14 .

The weighted sum does not follow a chi-squared distribution directly. The combined degrees of

freedom would be 11 + 14 = 25 , not 26 . This statement is false.

Option 3: The ratio $\frac{S_1^2}{S_2^2}$ is distributed as $F_{11,14}$ because it represents the ratio of two independent chi-squared

variables divided by their respective degrees of freedom. Multiplying by a constant (in this case, $\frac{5}{4}$)

does not change the F-distribution's form, just scales it. This statement is true.

Option 4: The mean of $\overline{Y}$ is -2 . The distribution of  $\overline{Y}$ is  $N(-2,\frac{5}{15})=N(-2,\frac{1}{3})$ . 

The quantity  $\overline{Y}$ - 2 shifts the mean to -4 . Scaling a normal distribution does not yield a t-distribution.

Instead,  $2\sqrt{3}(\overline{Y}-2)$ results in a normal distribution, not a t-distribution. This statement is false.

Hence, option 1) and 3) are correct.