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# The probability density function $p\left( x \right)$ cannot exceed:A. MeanB. ZeroC. OneD. Infinity

Last updated date: 30th May 2024
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Hint: Here, we will first use if $p\left( x \right)$ is a probability density function for a continuous random variable $X$ then $\int\limits_{ - \infty }^\infty {p\left( t \right)dt} = 1$. And then use the given conditions of the question, to find the required value.

Complete step-by-step solution:
Given that the function is a probability density function $p\left( x \right)$.

We know that a probability density function is a function of a continuous random variable, whose integral across an interval gives the probability that the value of the variable lies within the same interval.

So we have now seen that if $p\left( x \right)$ is the probability density function for a continuous random variable $X$ in any subset of the real numbers, then

$P\left( b \right) = P\left( {X \leqslant x} \right) \\ = \int\limits_{ - \infty }^x {p\left( t \right)dt} \\$

Since we also know that the probability density function must satisfy two conditions that it must be non-negative, so that the above integral is always non negative and it must integrate to one, that is, $p\left( x \right) \geqslant 0$ for any value of $x$ and $\int\limits_{ - \infty }^\infty {p\left( t \right)dt} = 1$, where the integral is implicitly taken over the whole real line

Thus, we can say from the above two conditions of the probability density function $p\left( x \right)$ that this function can never be greater than 1.

Therefore, the probability density function $p\left( x \right)$ cannot exceed one.

Hence, the option C is correct.

Note: In solving these types of questions, you should be familiar with the concept of probability function for a continuous random variable and its integration. Students should also remember the integration rules carefully for more accuracy. Then use the given conditions and values given in the question, to find the required values. Also, we are supposed to write the values properly to avoid any miscalculation.