### CSIR JUNE 2011 PART B QUESTION 52 SOLUTION (Random variable with $E(X) = Var(X)$)

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

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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.
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
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