Which of the following functions is not p.d.f. of a continuous random variable X?

Which of the following functions is not p.d.f. of a continuous random variable X?
  1. $F_{3}$
  2. $F_{4}$
  3. $F_{1}$
  4. $F_{2}$

Solution

Since $P(X=1)=P(X=2)=\ldots \ldots \ldots=P(X=n)$ and $P(X=1)+P(X=2)+\ldots \ldots+P(X=n)=1$, We get $P(X=1)=P(X=2)=P(X=n)=\frac{1}{n}$ \begin{array}{|c|c|c|c|} \hline \mathbf{x}_{\mathbf{i}} & \mathbf{p}_{\mathbf{i}} & \mathbf{x}_{\mathbf{i}} \mathbf{p}_{\mathbf{i}} & \mathbf{x}_{\mathbf{i}}^{2} \mathbf{p}_{\mathbf{i}} \\ \hline 1 & \frac{1}{\mathrm{n}} & \frac{1}{\mathrm{n}} & \frac{1}{\mathrm{n}} \\ 2 & \frac{1}{\mathrm{n}} & \frac{2}{\mathrm{n}} & \frac{2^{2}}{\mathrm{n}} \\ 3 & \frac{1}{\mathrm{n}} & \frac{3}{\mathrm{n}} & \frac{3^{2}}{\mathrm{n}} \\ \vdots & \vdots & \vdots & \vdots \\ \mathrm{n} & \frac{1}{\mathrm{n}} & \frac{\mathrm{n}}{\mathrm{n}} & \frac{\mathrm{n}^{2}}{\mathrm{n}} \\ \hline \end{array} $\sum \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{1}{\mathrm{n}}+\frac{2}{\mathrm{n}}+\frac{3}{\mathrm{n}}+\ldots \ldots+\frac{\mathrm{n}}{\mathrm{n}}=\frac{1}{\mathrm{n}}(1+2+3+\ldots \ldots \ldots \mathrm{n})=\frac{1}{\mathrm{n}} \frac{\mathrm{n}(\mathrm{n}+1)}{2}=\frac{\mathrm{n}+1}{2}$ $\begin{aligned} \sum x_{1}{ }^{2} p_{1} &=\frac{1^{2}}{n}+\frac{2^{2}}{n}+\frac{3^{2}}{n}+\ldots \ldots+\frac{n^{2}}{n} \\ &=\frac{1}{n}\left(1^{2}+2^{2}+3^{2}+\ldots+n^{2}\right)=\frac{1}{n} \frac{n(n+1)(2 n+1)}{6}-\frac{(n+1)(2 n+1)}{6} \end{aligned}$ Given $E(X)=V(X)$ $\begin{array}{l} \sum x_{i} p_{i}=\sum x_{i}^{2} p_{i}-\left[\sum x_{i} p_{i}\right]^{2} \\ \frac{n+1}{2}=\frac{(n+1)(2 n+1)}{6}-\left(\frac{n+1}{2}\right)^{2} \\ \frac{n+1}{2}=(n+1)\left(\frac{2 n+1}{6}-\frac{n+1}{4}\right) \Rightarrow \frac{1}{2}=\frac{8 n+4-6 n-6}{24} \\ 12=2 n-2 \Rightarrow n=7 \end{array}$

Asked in: MHT CET 2020 (12 Oct Shift 1)

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