Distribution MCQ Quiz - Objective Question with Answer for Distribution - Download Free PDF

Concept:

For an exponential distribution with parameter  $\lambda =1$ :
 

  $F(t)=1-e^{-t}$ is the CDF.

$f(t)=e^{-t}$ is the PDF.

Explanation:

Option 1:The maximum of two independent exponential random variables $X_1$ and $X_2$ follows the formula

$\mathbb{P}(max\{X_1,X_2\}<1)=F(1{)}^2=(1-e^{-1}{)}^2=(1-\frac{1}{e}{)}^2.$

and  $(1-\frac{1}{e}{)}^2=1-2\cdot \frac{1}{e}+\frac{1}{e^2}=1-\frac{2}{e}+\frac{1}{e^2}.$

This expression is not equal to $\frac{1}{2e}$  , so Option 1 is false.

Option 2: The minimum of two independent exponential random variables  $X_1$ and $X_2$ follows the formula:

$\mathbb{P}(min\{X_1,X_2\}>1)=(1-F(1){)}^2=e^{-1}\cdot e^{-1}=e^{-2}.$

This is not equal to $\frac{1}{2e}$ , so Option 2 is false.

Option 3: For three independent exponential random variables, we need to compute the probability that the minimum

of  $X_1$ and $X_2$ is greater than $X_3$ . The joint distribution and comparison of exponential variables leads to this result

being known as a standard property of exponential distributions. The probability for this scenario is $\frac{2}{3}$ , so Option 3 is true.

Option 4: Similarly, using properties of the exponential distribution and the comparison of order statistics, the probability

that the maximum of   $X_1$ and $X_2$   is less than $X_3$is known to be $\frac{1}{3}$, so Option 4 is true.

Hence, option 3) and 4) are correct.

Concept:

Probability Density Function (PDF) for Exponential Distribution

The probability density function (PDF) of an exponential distribution is given by:

$f(x|\theta )=\{\begin{array}{ll}\theta e^{-\theta x}& \text{if }x>0\\ 0& \text{otherwise}\end{array}$

Here, $\theta$ is the rate parameter. The exponential distribution is often used to model time

until the next event in processes where events occur continuously and independently

at a constant average rate (e.g., Poisson process).

From the sample $X_1,X_2,\dots ,X_n$, the sample mean $T_n=\frac{1}{n}\sum _{i=1}^n{X}_i$  is a sufficient statistic for $\theta$.

The sample mean $T_n$  is an estimator for $\frac{1}{\theta }$  because for an exponential distribution, the expected value is $\mathbb{E}(X_i)=\frac{1}{\theta }$.

Explanation: The problem you are dealing with involves a random sample $X_1,X_2,\dots ,X_n$ from a distribution with the given probability density function (PDF):

$f(x|\theta )=\{\begin{array}{ll}\theta e^{-\theta x}& \text{if }x>0\\ 0& \text{otherwise}\end{array}$

where $\theta >0$ is an unknown parameter, and the statistic $T_n=\frac{1}{n}\sum _{i=1}^n{X}_i$ is considered.

Option 1: For an exponential distribution, the sample mean $T_n=\frac{1}{n}\sum _{i=1}^n{X}_i$ is a sufficient and

unbiased estimator of $\frac{1}{\theta }$. Therefore, the unbiased estimator of $\theta$ is $\frac{1}{T_n}$.

The UMVUE of $\theta$ is $\frac{n-1}{n{T}_n}$, as this is a correction that adjusts for bias. So, option 1 is true.

 Option 2: The Fisher information $I(\theta )$ for a single observation from the exponential distribution $\theta e^{-\theta x}$ is $\frac{1}{{\theta }^2}$.

Therefore, for a sample of size n, the Fisher information is $n\times \frac{1}{{\theta }^2}$ , and the Cramer-Rao lower bound

(CRLB) for the variance of any unbiased estimator of $\theta$ is:

$\text{CRLB}=\frac{1}{n\times I(\theta )}=\frac{{\theta }^2}{n}.$

So, option 2 is true.

Option 3: The UMVUE of $\theta$ does not necessarily attain the Cramer-Rao lower bound. While it is unbiased

and has the minimum variance among unbiased estimators, it might not reach the bound. Option 3 is false.

Option 4: The probability $P_{\theta }(X_1\le 1)$ can be calculated from the cumulative distribution function

(CDF) of the exponential distribution:

$P_{\theta }(X_1\le 1)=1-e^{-\theta }.$

Using the fact that $T_n$ is an unbiased estimator for $\frac{1}{\theta }$, the expression $(1-e^{-\frac{1}{T_n}})$ is a consistent

estimator for $1-e^{-\theta }$, which is $P_{\theta }(X_1\le 1)$. Therefore, option 4 is true.

The correct statements are Option 1), Option 2) and Option 4).

Concept:

Expected Value Calculation:

The expected value $E(|Z_1-Z_2|)$  can be derived using geometric probability or integrating over

the length of the segment. For a uniform distribution of the point P on the segment, the expected value of

the absolute difference between the two distances is known to be

$E(|Z_1-Z_2|)=\frac{\alpha }{2}$

Explanation:

Let us introduce the variable $X=Z_1-Z_2$. Since $Z_1+Z_2=\alpha$, we have  $Z_2=\alpha -Z_1$

Thus, $X=Z_1-(\alpha -Z_1)=2Z_1-\alpha$ , so  $X=2Z_1-\alpha$ .

We want to compute $E(|X|)$, which is equivalent to $E(|2Z_1-\alpha |$).

Distribution of  $Z_1$: 

Since  P  is chosen uniformly along the line segment $Z_1$, is uniformly distributed

between  0 and  α. The probability density function (PDF) of $Z_1$ is

  $f_{Z_1}(z_1)=\frac{1}{\alpha }\text{for}0\le z_1\le \alpha$

Expected Value of $|X|$: 

Now, we compute the expected value $E(|2Z_1-\alpha |)$. This involves integrating the

absolute value of $2Z_1-\alpha$ over the range of $Z_1$:

$E(|2Z_1-\alpha |)={\int }_0^{\alpha }|2z_1-\alpha |\cdot \frac{1}{\alpha }d{z}_1$

Splitting the Integral

The expression $|2Z_1-\alpha |$ changes sign when $z_1=\frac{\alpha }{2}$.

So, we split the integral into two parts:

For $0\le z_1\le \frac{\alpha }{2}$, $|2z_1-\alpha |=\alpha -2z_1$.

For $\frac{\alpha }{2}\le z_1\le \alpha$, $|2z_1-\alpha |=2z_1-\alpha$.

Therefore, the expected value becomes:

 $E(|2Z_1-\alpha |)={\int }_0^{\frac{\alpha }{2}}(\alpha -2z_1)\cdot \frac{1}{\alpha }d{z}_1+{\int }_{\frac{\alpha }{2}}^{\alpha }(2z_1-\alpha )\cdot \frac{1}{\alpha }d{z}_1$   

First integral ${\int }_0^{\frac{\alpha }{2}}(\alpha -2z_1)\cdot \frac{1}{\alpha }d{z}_1=\frac{1}{\alpha }{[\alpha z_1-z_1^2]}_0^{\frac{\alpha }{2}}=\frac{1}{\alpha }(\frac{{\alpha }^2}{2}-\frac{{\alpha }^2}{4})=\frac{\alpha }{4}$

Second integral ${\int }_{\frac{\alpha }{2}}^{\alpha }(2z_1-\alpha )\cdot \frac{1}{\alpha }d{z}_1=\frac{1}{\alpha }{[z_1^2-\alpha z_1]}_{\frac{\alpha }{2}}^{\alpha }=\frac{1}{\alpha }({\alpha }^2-{\alpha }^2+\frac{{\alpha }^2}{4})=\frac{\alpha }{4}$
 
 $E(|2Z_1-\alpha |)=\frac{\alpha }{4}+\frac{\alpha }{4}=\frac{\alpha }{2}$
 
$E(|Z_1-Z_2|)=\frac{\alpha }{2}$

Hence option 3) is correct.
 

The correct answer is options are 1, 2 & 4. 

we will update the solution as soon as possible.

The correct answer is options are 2 & 4. 

we will update the solution as soon as possible.

Concept:

Expected Value Calculation:

The expected value $E(|Z_1-Z_2|)$  can be derived using geometric probability or integrating over

the length of the segment. For a uniform distribution of the point P on the segment, the expected value of

the absolute difference between the two distances is known to be

$E(|Z_1-Z_2|)=\frac{\alpha }{2}$

Explanation:

Let us introduce the variable $X=Z_1-Z_2$. Since $Z_1+Z_2=\alpha$, we have  $Z_2=\alpha -Z_1$

Thus, $X=Z_1-(\alpha -Z_1)=2Z_1-\alpha$ , so  $X=2Z_1-\alpha$ .

We want to compute $E(|X|)$, which is equivalent to $E(|2Z_1-\alpha |$).

Distribution of  $Z_1$: 

Since  P  is chosen uniformly along the line segment $Z_1$, is uniformly distributed

between  0 and  α. The probability density function (PDF) of $Z_1$ is

  $f_{Z_1}(z_1)=\frac{1}{\alpha }\text{for}0\le z_1\le \alpha$

Expected Value of $|X|$: 

Now, we compute the expected value $E(|2Z_1-\alpha |)$. This involves integrating the

absolute value of $2Z_1-\alpha$ over the range of $Z_1$:

$E(|2Z_1-\alpha |)={\int }_0^{\alpha }|2z_1-\alpha |\cdot \frac{1}{\alpha }d{z}_1$

Splitting the Integral

The expression $|2Z_1-\alpha |$ changes sign when $z_1=\frac{\alpha }{2}$.

So, we split the integral into two parts:

For $0\le z_1\le \frac{\alpha }{2}$, $|2z_1-\alpha |=\alpha -2z_1$.

For $\frac{\alpha }{2}\le z_1\le \alpha$, $|2z_1-\alpha |=2z_1-\alpha$.

Therefore, the expected value becomes:

 $E(|2Z_1-\alpha |)={\int }_0^{\frac{\alpha }{2}}(\alpha -2z_1)\cdot \frac{1}{\alpha }d{z}_1+{\int }_{\frac{\alpha }{2}}^{\alpha }(2z_1-\alpha )\cdot \frac{1}{\alpha }d{z}_1$   

First integral ${\int }_0^{\frac{\alpha }{2}}(\alpha -2z_1)\cdot \frac{1}{\alpha }d{z}_1=\frac{1}{\alpha }{[\alpha z_1-z_1^2]}_0^{\frac{\alpha }{2}}=\frac{1}{\alpha }(\frac{{\alpha }^2}{2}-\frac{{\alpha }^2}{4})=\frac{\alpha }{4}$

Second integral ${\int }_{\frac{\alpha }{2}}^{\alpha }(2z_1-\alpha )\cdot \frac{1}{\alpha }d{z}_1=\frac{1}{\alpha }{[z_1^2-\alpha z_1]}_{\frac{\alpha }{2}}^{\alpha }=\frac{1}{\alpha }({\alpha }^2-{\alpha }^2+\frac{{\alpha }^2}{4})=\frac{\alpha }{4}$
 
 $E(|2Z_1-\alpha |)=\frac{\alpha }{4}+\frac{\alpha }{4}=\frac{\alpha }{2}$
 
$E(|Z_1-Z_2|)=\frac{\alpha }{2}$

Hence option 3) is correct.
 

Given:-

X1, X2, ..., Xn are i.i.d. random variables with common pdf

$f(x\mid \theta )=\frac{(\log \theta ){\theta }^x}{\theta -1}$

Concept Used:-

To check whether T = ∑ᵢ Xᵢ is a sufficient statistic for the parameter θ, we need to find the likelihood function of the sample.

Explanation:-

The joint pdf of the sample is given by

$$\begin{gathered} \Rightarrow f(x₁,x₂,...,xₙ|\theta )=f(x₁|\theta )f(x₂|\theta )...f(xₙ|\theta ) \\ \Rightarrow f(x₁,x₂,...,xₙ|\theta )=\prod _{i=1}^{n}f(x_i\mid \theta ) \\ \Rightarrow f(x₁,x₂,...,xₙ|\theta )=\prod _{i=1}^n\frac{(\log \theta ){\theta }^{x_i}}{\theta -1} \\ \Rightarrow f(x₁,x₂,...,xₙ|\theta )=(\log \theta {)}^n\frac{{\theta }^{\sum _{i=1}^n{x}_i}}{(\theta -1{)}^n} \end{gathered}$$

The factorization theorem states that a statistic T is sufficient for a parameter θ if and only if the joint pdf of the sample can be factorized into two functions.

One which depends on the sample only through T, and the other that does not depend on θ.

We can rewrite the joint pdf of the sample as

⇒$f(x₁,x₂,...,xₙ|\theta )=h(x₁,x₂,...,xₙ)g(T|\theta )$,

Here,

$$\begin{gathered} \Rightarrow h(x₁,x₂,...,xₙ)=(\log \theta {)}^n\text{and} \\ \Rightarrow g(T|\theta )=({\theta }^T)/(\theta -1{)}^n \end{gathered}$$  

Since the joint pdf can be factorized in this way, T is a sufficient statistic for θ.

To check whether T is complete, we need to show that T is able to capture all the information about θ contained in the sample.

In other words, any function of T that is independent of θ must have an expected value equal to 0.

Let h(T) be a function of T that is independent of θ. Then

$$\begin{gathered} E[h(T)]={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}h(T)g(T|\theta )dT \\ E[h(T)]={\int }_0^{1}h(T)\frac{{\theta }^T}{(\theta -1{)}^n}dT \\ E[h(T)]=\frac{(\log \theta {)}^n}{(\theta -1{)}^n}{\int }_0^{1}h(T)\exp (T\log \theta )dT \end{gathered}$$

Since θ > 1, \log θ > 0, and the integrand is positive on [0,1].

Therefore, if E[h(T)] = 0 for all θ > 1, then h(T) must be identically zero almost everywhere on [0,1].

This shows that T is a complete, sufficient statistic for θ.

Therefore, the statistic $T=\sum _{i=1}^n{X}_i$ is completely sufficient.

Hence, the correct option is 3.

The correct answer is option 2

we will update the solution as soon as possible.

Concept:

Expected Value Calculation:

The expected value $E(|Z_1-Z_2|)$  can be derived using geometric probability or integrating over

the length of the segment. For a uniform distribution of the point P on the segment, the expected value of

the absolute difference between the two distances is known to be

$E(|Z_1-Z_2|)=\frac{\alpha }{2}$

Explanation:

Let us introduce the variable $X=Z_1-Z_2$. Since $Z_1+Z_2=\alpha$, we have  $Z_2=\alpha -Z_1$

Thus, $X=Z_1-(\alpha -Z_1)=2Z_1-\alpha$ , so  $X=2Z_1-\alpha$ .

We want to compute $E(|X|)$, which is equivalent to $E(|2Z_1-\alpha |$).

Distribution of  $Z_1$: 

Since  P  is chosen uniformly along the line segment $Z_1$, is uniformly distributed

between  0 and  α. The probability density function (PDF) of $Z_1$ is

  $f_{Z_1}(z_1)=\frac{1}{\alpha }\text{for}0\le z_1\le \alpha$

Expected Value of $|X|$: 

Now, we compute the expected value $E(|2Z_1-\alpha |)$. This involves integrating the

absolute value of $2Z_1-\alpha$ over the range of $Z_1$:

$E(|2Z_1-\alpha |)={\int }_0^{\alpha }|2z_1-\alpha |\cdot \frac{1}{\alpha }d{z}_1$

Splitting the Integral

The expression $|2Z_1-\alpha |$ changes sign when $z_1=\frac{\alpha }{2}$.

So, we split the integral into two parts:

For $0\le z_1\le \frac{\alpha }{2}$, $|2z_1-\alpha |=\alpha -2z_1$.

For $\frac{\alpha }{2}\le z_1\le \alpha$, $|2z_1-\alpha |=2z_1-\alpha$.

Therefore, the expected value becomes:

 $E(|2Z_1-\alpha |)={\int }_0^{\frac{\alpha }{2}}(\alpha -2z_1)\cdot \frac{1}{\alpha }d{z}_1+{\int }_{\frac{\alpha }{2}}^{\alpha }(2z_1-\alpha )\cdot \frac{1}{\alpha }d{z}_1$   

First integral ${\int }_0^{\frac{\alpha }{2}}(\alpha -2z_1)\cdot \frac{1}{\alpha }d{z}_1=\frac{1}{\alpha }{[\alpha z_1-z_1^2]}_0^{\frac{\alpha }{2}}=\frac{1}{\alpha }(\frac{{\alpha }^2}{2}-\frac{{\alpha }^2}{4})=\frac{\alpha }{4}$

Second integral ${\int }_{\frac{\alpha }{2}}^{\alpha }(2z_1-\alpha )\cdot \frac{1}{\alpha }d{z}_1=\frac{1}{\alpha }{[z_1^2-\alpha z_1]}_{\frac{\alpha }{2}}^{\alpha }=\frac{1}{\alpha }({\alpha }^2-{\alpha }^2+\frac{{\alpha }^2}{4})=\frac{\alpha }{4}$
 
 $E(|2Z_1-\alpha |)=\frac{\alpha }{4}+\frac{\alpha }{4}=\frac{\alpha }{2}$
 
$E(|Z_1-Z_2|)=\frac{\alpha }{2}$

Hence option 3) is correct.
 

The correct answer is options are 1, 2 & 4. 

we will update the solution as soon as possible.

Given:-

X1, X2, ..., Xn are i.i.d. random variables with common pdf

$f(x\mid \theta )=\frac{(\log \theta ){\theta }^x}{\theta -1}$

Concept Used:-

To check whether T = ∑ᵢ Xᵢ is a sufficient statistic for the parameter θ, we need to find the likelihood function of the sample.

Explanation:-

The joint pdf of the sample is given by

$$\begin{gathered} \Rightarrow f(x₁,x₂,...,xₙ|\theta )=f(x₁|\theta )f(x₂|\theta )...f(xₙ|\theta ) \\ \Rightarrow f(x₁,x₂,...,xₙ|\theta )=\prod _{i=1}^{n}f(x_i\mid \theta ) \\ \Rightarrow f(x₁,x₂,...,xₙ|\theta )=\prod _{i=1}^n\frac{(\log \theta ){\theta }^{x_i}}{\theta -1} \\ \Rightarrow f(x₁,x₂,...,xₙ|\theta )=(\log \theta {)}^n\frac{{\theta }^{\sum _{i=1}^n{x}_i}}{(\theta -1{)}^n} \end{gathered}$$

The factorization theorem states that a statistic T is sufficient for a parameter θ if and only if the joint pdf of the sample can be factorized into two functions.

One which depends on the sample only through T, and the other that does not depend on θ.

We can rewrite the joint pdf of the sample as

⇒$f(x₁,x₂,...,xₙ|\theta )=h(x₁,x₂,...,xₙ)g(T|\theta )$,

Here,

$$\begin{gathered} \Rightarrow h(x₁,x₂,...,xₙ)=(\log \theta {)}^n\text{and} \\ \Rightarrow g(T|\theta )=({\theta }^T)/(\theta -1{)}^n \end{gathered}$$  

Since the joint pdf can be factorized in this way, T is a sufficient statistic for θ.

To check whether T is complete, we need to show that T is able to capture all the information about θ contained in the sample.

In other words, any function of T that is independent of θ must have an expected value equal to 0.

Let h(T) be a function of T that is independent of θ. Then

$$\begin{gathered} E[h(T)]={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}h(T)g(T|\theta )dT \\ E[h(T)]={\int }_0^{1}h(T)\frac{{\theta }^T}{(\theta -1{)}^n}dT \\ E[h(T)]=\frac{(\log \theta {)}^n}{(\theta -1{)}^n}{\int }_0^{1}h(T)\exp (T\log \theta )dT \end{gathered}$$

Since θ > 1, \log θ > 0, and the integrand is positive on [0,1].

Therefore, if E[h(T)] = 0 for all θ > 1, then h(T) must be identically zero almost everywhere on [0,1].

This shows that T is a complete, sufficient statistic for θ.

Therefore, the statistic $T=\sum _{i=1}^n{X}_i$ is completely sufficient.

Hence, the correct option is 3.

Concept:

For an exponential distribution with parameter  $\lambda =1$ :
 

  $F(t)=1-e^{-t}$ is the CDF.

$f(t)=e^{-t}$ is the PDF.

Explanation:

Option 1:The maximum of two independent exponential random variables $X_1$ and $X_2$ follows the formula

$\mathbb{P}(max\{X_1,X_2\}<1)=F(1{)}^2=(1-e^{-1}{)}^2=(1-\frac{1}{e}{)}^2.$

and  $(1-\frac{1}{e}{)}^2=1-2\cdot \frac{1}{e}+\frac{1}{e^2}=1-\frac{2}{e}+\frac{1}{e^2}.$

This expression is not equal to $\frac{1}{2e}$  , so Option 1 is false.

Option 2: The minimum of two independent exponential random variables  $X_1$ and $X_2$ follows the formula:

$\mathbb{P}(min\{X_1,X_2\}>1)=(1-F(1){)}^2=e^{-1}\cdot e^{-1}=e^{-2}.$

This is not equal to $\frac{1}{2e}$ , so Option 2 is false.

Option 3: For three independent exponential random variables, we need to compute the probability that the minimum

of  $X_1$ and $X_2$ is greater than $X_3$ . The joint distribution and comparison of exponential variables leads to this result

being known as a standard property of exponential distributions. The probability for this scenario is $\frac{2}{3}$ , so Option 3 is true.

Option 4: Similarly, using properties of the exponential distribution and the comparison of order statistics, the probability

that the maximum of   $X_1$ and $X_2$   is less than $X_3$is known to be $\frac{1}{3}$, so Option 4 is true.

Hence, option 3) and 4) are correct.

Concept:

Probability Density Function (PDF) for Exponential Distribution

The probability density function (PDF) of an exponential distribution is given by:

$f(x|\theta )=\{\begin{array}{ll}\theta e^{-\theta x}& \text{if }x>0\\ 0& \text{otherwise}\end{array}$

Here, $\theta$ is the rate parameter. The exponential distribution is often used to model time

until the next event in processes where events occur continuously and independently

at a constant average rate (e.g., Poisson process).

From the sample $X_1,X_2,\dots ,X_n$, the sample mean $T_n=\frac{1}{n}\sum _{i=1}^n{X}_i$  is a sufficient statistic for $\theta$.

The sample mean $T_n$  is an estimator for $\frac{1}{\theta }$  because for an exponential distribution, the expected value is $\mathbb{E}(X_i)=\frac{1}{\theta }$.

Explanation: The problem you are dealing with involves a random sample $X_1,X_2,\dots ,X_n$ from a distribution with the given probability density function (PDF):

$f(x|\theta )=\{\begin{array}{ll}\theta e^{-\theta x}& \text{if }x>0\\ 0& \text{otherwise}\end{array}$

where $\theta >0$ is an unknown parameter, and the statistic $T_n=\frac{1}{n}\sum _{i=1}^n{X}_i$ is considered.

Option 1: For an exponential distribution, the sample mean $T_n=\frac{1}{n}\sum _{i=1}^n{X}_i$ is a sufficient and

unbiased estimator of $\frac{1}{\theta }$. Therefore, the unbiased estimator of $\theta$ is $\frac{1}{T_n}$.

The UMVUE of $\theta$ is $\frac{n-1}{n{T}_n}$, as this is a correction that adjusts for bias. So, option 1 is true.

 Option 2: The Fisher information $I(\theta )$ for a single observation from the exponential distribution $\theta e^{-\theta x}$ is $\frac{1}{{\theta }^2}$.

Therefore, for a sample of size n, the Fisher information is $n\times \frac{1}{{\theta }^2}$ , and the Cramer-Rao lower bound

(CRLB) for the variance of any unbiased estimator of $\theta$ is:

$\text{CRLB}=\frac{1}{n\times I(\theta )}=\frac{{\theta }^2}{n}.$

So, option 2 is true.

Option 3: The UMVUE of $\theta$ does not necessarily attain the Cramer-Rao lower bound. While it is unbiased

and has the minimum variance among unbiased estimators, it might not reach the bound. Option 3 is false.

Option 4: The probability $P_{\theta }(X_1\le 1)$ can be calculated from the cumulative distribution function

(CDF) of the exponential distribution:

$P_{\theta }(X_1\le 1)=1-e^{-\theta }.$

Using the fact that $T_n$ is an unbiased estimator for $\frac{1}{\theta }$, the expression $(1-e^{-\frac{1}{T_n}})$ is a consistent

estimator for $1-e^{-\theta }$, which is $P_{\theta }(X_1\le 1)$. Therefore, option 4 is true.

The correct statements are Option 1), Option 2) and Option 4).

The correct answer is options are 2 & 4. 

we will update the solution as soon as possible.