Let $X$ be a random variable with $E(x) = Var(X)$. Then the distribution of $X$ has to be

1. Poisson,

2. Exponential,

3. Normal,

4. cannot be identified from the given condition

Poisson: Let $X$ be a discrete random variable. $X$ has a Poisson distribution with a positive real parameter $\lambda$ if the probability distribution function of $X$ is equal to $$\frac{\lambda^n e^{-\lambda}}{n!}$$ for $n \ge 0$. In this case mean = variance = $\lambda$.

Normal: Let $X$ be a continuous random variable. $X$ has a Normal distribution with parameters $\mu$ and $\sigma$ if the probability distribution of $X$ is equal to $$\frac{1}{\sigma \sqrt{2\pi}}e^{\frac{-1}{2}(\frac{x-\mu}{\sigma})^2}.$$ In this case mean = $\mu$ and variance = $\sigma^2$. It is possible to take $\mu = 1$ and $\sigma = 1$ the we get $mean = variance$.

So we conclude that the condition $E(X) = Var(X)$ can occur for both Poisson and Normal. Also, for many other distributions and please comment below if you know any other such distribution. This shows that from the condition $E(X) = Var(X)$ we cannot conclude the distribution uniquely.

1. Poisson,

2. Exponential,

3. Normal,

4. cannot be identified from the given condition

**I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.****Solution:**Poisson: Let $X$ be a discrete random variable. $X$ has a Poisson distribution with a positive real parameter $\lambda$ if the probability distribution function of $X$ is equal to $$\frac{\lambda^n e^{-\lambda}}{n!}$$ for $n \ge 0$. In this case mean = variance = $\lambda$.

Normal: Let $X$ be a continuous random variable. $X$ has a Normal distribution with parameters $\mu$ and $\sigma$ if the probability distribution of $X$ is equal to $$\frac{1}{\sigma \sqrt{2\pi}}e^{\frac{-1}{2}(\frac{x-\mu}{\sigma})^2}.$$ In this case mean = $\mu$ and variance = $\sigma^2$. It is possible to take $\mu = 1$ and $\sigma = 1$ the we get $mean = variance$.

So we conclude that the condition $E(X) = Var(X)$ can occur for both Poisson and Normal. Also, for many other distributions and please comment below if you know any other such distribution. This shows that from the condition $E(X) = Var(X)$ we cannot conclude the distribution uniquely.

**so option 4 is correct**.**Click here for more problems.**

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